Binomial Theorem

You already know how to square a binomial: $(a+b)^2 = a^2 + 2ab + b^2$ . You may even remember the cube, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ , useful for evaluating things like $(98)^2 = (100-2)^2$ without brute-force multiplication. But what about $(98)^5$ , or $(101)^6$ , or the general $(a+b)^n$ for any positive integer $n$ ? Multiplying $(a+b)$ by itself fifty times is not a plan — it is a punishment. This chapter hands you the escape route: the Binomial Theorem, a single formula that expands $(a+b)^n$ for every positive integer $n$ in one stroke, without ever performing a repeated multiplication. The theorem is not pulled from nowhere. It grows directly out of a pattern you can literally draw — Pascal's triangle — and out of the counting formula $^{n}C_r$ you met in the previous chapter on permutations and combinations. Every coefficient in the expansion of $(a+b)^n$ turns out to be exactly $^{n}C_r$ , the number of ways to choose $r$ objects from $n$ ; the algebra of expanding a power and the combinatorics of choosing subsets are, astonishingly, the very same numbers. Once proved by mathematical induction, the theorem unlocks tools used constantly through the rest of school and competitive mathematics: a general term formula that plucks out any single term of an expansion — the $100$ th, say — without writing the other ninety-nine; a rule for the middle term (or terms); techniques for finding a term independent of a variable or the coefficient of a particular power; and special cases such as $(1+x)^n$ that generate the identity $2^n = {}^{n}C_0 + {}^{n}C_1 + \cdots + {}^{n}C_n$ . This article walks the full chapter: the pattern in Pascal's triangle, the statement and induction proof of the theorem, the general term and how to hunt for a specific term, the middle-term rule with a rigorous derivation of where it sits, the special expansions of $(x-y)^n$ and $(1+x)^n$ , and fully worked board-level and JEE/NEET-level problems — including the classic "term independent of $x$ " question type that appears on nearly every competitive paper touching this chapter.

Key takeaways

Binomial Theorem (positive integral index): $(a+b)^n = \sum_{k=0}^{n} {}^{n}C_k\, a^{n-k}b^k$ , valid for every positive integer $n$ ; it has exactly $n+1$ terms.
The coefficients $^{n}C_0, {}^{n}C_1, \ldots, {}^{n}C_n$ are the binomial coefficients; arranged row by row for $n = 0, 1, 2, \ldots$ they form Pascal's triangle, where every interior entry is the sum of the two entries above it.
In every term of the expansion, the powers of $a$ and $b$ always add up to $n$ ; the power of $a$ decreases from $n$ to $0$ while the power of $b$ increases from $0$ to $n$ .
The general term (the $(r+1)$ th term, not the $r$ th) is $T_{r+1} = {}^{n}C_r\, a^{n-r}b^r$ — the single most-used formula in the chapter, since it lets you locate any specific term or its coefficient directly.
For even $n$ there is one middle term, the $\left(\tfrac{n}{2}+1\right)$ th term; for odd $n$ there are two middle terms, the $\left(\tfrac{n+1}{2}\right)$ th and $\left(\tfrac{n+3}{2}\right)$ th, and these carry the greatest binomial coefficients.
Special cases fall straight out of the theorem: $(x-y)^n = \sum_{k=0}^n (-1)^k\,{}^{n}C_k\,x^{n-k}y^k$ (signs alternate), and $(1+x)^n = \sum_{k=0}^n {}^{n}C_k\,x^k$ .
Putting $x=1$ in $(1+x)^n$ gives $2^n = {}^{n}C_0 + {}^{n}C_1 + \cdots + {}^{n}C_n$ — the sum of all binomial coefficients in a row of Pascal's triangle; putting $x=-1$ gives the alternating sum $0 = {}^{n}C_0 - {}^{n}C_1 + {}^{n}C_2 - \cdots$ .
The theorem is proved for all positive integers $n$ by the principle of mathematical induction, using the Pascal identity $^{k}C_r + {}^{k}C_{r-1} = {}^{k+1}C_r$ to pass from $n=k$ to $n=k+1$ .
Splitting an awkward number as a sum or difference of two easy numbers (e.g. $98 = 100-2$ ) and expanding by the Binomial Theorem turns tedious powers like $(98)^5$ into a short, exact arithmetic computation.
Key idea: (a+b)^n expands into exactly n+1 terms whose coefficients are the binomial coefficients nCr of Pascal's triangle, and every term is captured in one formula, the general term T(r+1) = nCr a^(n-r) b^r, from which the middle term, a specific coefficient, and identities like the sum 2^n of a row's coefficients all follow.

The core insight

The Binomial Theorem states $(a+b)^n = \sum_{k=0}^{n} {}^{n}C_k\, a^{n-k}b^k$ for every positive integer $n$ , so every one of its $n+1$ terms is captured by the single general-term formula $T_{r+1} = {}^{n}C_r\, a^{n-r}b^r$ , from which the middle term (or terms), a chosen coefficient, or the identity $2^n = \sum_r {}^{n}C_r$ can all be read off directly, without ever expanding the whole product by repeated multiplication.

What this chapter covers

The chapter opens by revisiting the familiar expansions of $(a+b)^0$ through $(a+b)^4$ and extracting three visual patterns: the number of terms, the rise-and-fall of the powers of $a$ and $b$ , and the constant sum of exponents. These patterns are then organised into Pascal's triangle, first as a triangle of raw numbers built by simple addition, and then rewritten using the combination symbol $^{n}C_r$ once the connection to counting is noticed. That rewriting is the launching pad for the Binomial Theorem itself, stated for any positive integral index $n$ and proved rigorously by the principle of mathematical induction. From the theorem the chapter derives its two most exam-tested tools: the general term $T_{r+1}$ , used to locate or identify any single term of an expansion, and the middle term rule, which identifies the term (or pair of terms) carrying the largest binomial coefficient. The chapter closes with the special cases $(x-y)^n$ and $(1+x)^n$ , the identity $2^n = \sum {}^{n}C_r$ that falls out of $(1+x)^n$ at $x=1$ , and a battery of applications: expanding algebraic expressions, computing large powers like $(98)^5$ by splitting them into convenient sums, comparing the size of expressions like $(1.01)^{1000000}$ against $10{,}000$ , and proving divisibility results such as " $6^n - 5^n$ leaves remainder $1$ on division by $25$ ."

The core ideas

Pascal's triangle: the pattern behind the coefficients

Look again at the expansions you already know.

(a+b)^0 = 1, \quad (a+b)^1 = a+b, \quad (a+b)^2 = a^2+2ab+b^2, \quad (a+b)^3 = a^3+3a^2b+3ab^2+b^3. \qquad (1)

where: each expansion is written out in full so the pattern is visible by inspection, not proved yet. Three observations recur every time. First, the number of terms in the expansion of $(a+b)^n$ is always $n+1$ , one more than the index — three terms for index $2$ , four terms for index $3$ . Second, in successive terms the power of $a$ steadily falls from $n$ down to $0$ while the power of $b$ steadily climbs from $0$ up to $n$ . Third, in every single term the two exponents add up to exactly $n$ , the index of the binomial — a built-in checksum you can use to catch algebra slips instantly. Writing only the numerical coefficients of each row, one below the next, produces a triangular array in which $1$ sits at the very top and runs down both slanting edges, and every interior number is the sum of the two numbers immediately above it (with a blank counted as $0$ ). This is Pascal's triangle, named for the French mathematician Blaise Pascal, though the same array was known centuries earlier in India as the Meru Prastara, described by the mathematician Pingala. Once a row is known, the next row is built purely by addition, with no multiplication required — the row for index $5$ , namely $1, 5, 10, 10, 5, 1$ , lets you write $(2x+3y)^5$ directly, term by term, without touching indices $0$ through $4$ again.

Figure

Pascals triangle from row zero to row six with each entry equal to the sum of the two entries above it and the largest entry or entries highlighted in red at the centre of every row

The Binomial Theorem for a positive integral index

Replace the numbers of Pascal's triangle with their combinatorial names. Row $n$ of the triangle is exactly $^{n}C_0, {}^{n}C_1, {}^{n}C_2, \ldots, {}^{n}C_n$ , and substituting these into the pattern observed above gives the theorem in full generality.

(a+b)^n = {}^{n}C_0\,a^n + {}^{n}C_1\,a^{n-1}b + {}^{n}C_2\,a^{n-2}b^2 + \cdots + {}^{n}C_{n-1}\,a\,b^{n-1} + {}^{n}C_n\,b^n = \sum_{k=0}^{n} {}^{n}C_k\, a^{n-k}b^k. \qquad (2)

where: $n$ is any positive integer, $a$ and $b$ are any real (or complex) numbers, and $^{n}C_k = \dfrac{n!}{k!\,(n-k)!}$ for $0 \le k \le n$ is the binomial coefficient — precisely the number of ways to choose $k$ objects out of $n$ , carried over unchanged from the previous chapter on combinations. The summation notation on the right is only shorthand for the fully written-out sum on the left; both say the same thing. Every term has the form (binomial coefficient) times (decreasing power of $a$ ) times (increasing power of $b$ ), and because $^{n}C_0 = {}^{n}C_n = 1$ always, the expansion begins with $a^n$ and ends with $b^n$ , each with coefficient exactly $1$ — matching the " $1$ " that sits at both ends of every row of Pascal's triangle.

The general term: naming any single term of the expansion

Writing out an entire expansion just to read off, say, its sixth term is wasteful. The summation form of the theorem hands you a direct formula for any individual term, called the general term.

T_{r+1} = {}^{n}C_r\, a^{n-r}\,b^r, \qquad r = 0, 1, 2, \ldots, n. \qquad (3)

where: $T_{r+1}$ denotes the term occupying position $r+1$ counted from the start of the expansion, so $r=0$ gives the first term $T_1 = {}^{n}C_0 a^n$ , $r=1$ gives the second term $T_2 = {}^{n}C_1 a^{n-1}b$ , and so on up to $r=n$ giving the last term $T_{n+1} = {}^{n}C_n b^n$ . The offset by one is the single most common source of error in this chapter: the term that carries $b^r$ is the $(r+1)$ th term, never the $r$ th. To find, for instance, the $7$ th term of an expansion you set $r+1=7$ , i.e. $r=6$ , and substitute into $(3)$ — you never set $r=7$ directly. This formula is the workhorse of the chapter: every "find the coefficient of $x^m$ ," "find the term independent of $x$ ," and "find the $k$ th term" question is solved by writing down $T_{r+1}$ for the given binomial, simplifying the power of the variable, and solving an equation for $r$ .

Figure

Bar chart of the binomial coefficients 8 choose r for r running from 0 to 8 forming a symmetric bell shape that peaks at the single middle term where r equals 4

The middle term or terms

Because there are always $n+1$ terms in the expansion of $(a+b)^n$ , the position of the "middle" depends on whether $n+1$ is odd or even — which depends, in turn, on whether $n$ itself is even or odd.

n \text{ even} \Rightarrow \text{one middle term: } T_{\frac{n}{2}+1}. \qquad n \text{ odd} \Rightarrow \text{two middle terms: } T_{\frac{n+1}{2}} \text{ and } T_{\frac{n+3}{2}}. \qquad (4)

where: for even $n$ , $n+1$ is odd, so the $n+1$ terms have a single central one, landing exactly halfway — the $\left(\frac{n}{2}+1\right)$ th term. For odd $n$ , $n+1$ is even, so there is no single centre; the two terms flanking the middle, at positions $\frac{n+1}{2}$ and $\frac{n+3}{2}$ , share the honour. For example, $(a+b)^{10}$ has $11$ terms, so its unique middle term is $T_6$ (setting $n=10$ : $\frac{10}{2}+1 = 6$ ); while $(a+b)^{9}$ has $10$ terms, so its two middle terms are $T_5$ and $T_6$ (setting $n=9$ : $\frac{9+1}{2}=5$ and $\frac{9+3}{2}=6$ ). The second derivation below proves that these middle positions are exactly where the largest binomial coefficient(s) occur — visible already in the reds highlighted in the Pascal's-triangle figure and in the peak of the bar chart above.

Special expansions and the sum of binomial coefficients

Three substitutions into the Binomial Theorem generate results used constantly elsewhere in the syllabus. Replacing $b$ by $-b$ flips the sign of every alternate term; setting $a=1, b=x$ produces the polynomial form most useful in algebra; and evaluating that polynomial form at $x=1$ produces a famous counting identity.

(x-y)^n = \sum_{k=0}^{n} (-1)^k\, {}^{n}C_k\, x^{n-k}y^k, \qquad (1+x)^n = \sum_{k=0}^{n} {}^{n}C_k\, x^k, \qquad 2^n = \sum_{k=0}^{n} {}^{n}C_k. \qquad (5)

where: in the first identity the sign alternates term by term because $(-y)^k = (-1)^k y^k$ , so terms with even $k$ stay positive and terms with odd $k$ turn negative — a frequent source of sign slips if the alternation is forgotten. The second identity is simply the Binomial Theorem with $a$ set to $1$ , so $a^{n-k}=1$ vanishes from every term and only the powers of $x$ survive; this is the form usually meant when a problem says "expand $(1+x)^n$ ." The third identity comes from setting $x=1$ in the second: $(1+1)^n = 2^n$ equals the sum of every binomial coefficient in row $n$ of Pascal's triangle — a clean way to see why the entries of each row of the triangle always sum to a power of $2$ . A companion identity, obtained by setting $x=-1$ instead, gives $0 = {}^{n}C_0 - {}^{n}C_1 + {}^{n}C_2 - \cdots + (-1)^n\,{}^{n}C_n$ : the alternating sum of any row of Pascal's triangle (for $n \ge 1$ ) is always exactly zero.

Two proofs and a locating rule

Derivation AProof of the Binomial Theorem by mathematical induction

The Binomial Theorem is a claim about every positive integer $n$ at once, and no finite amount of checking cases like $n=1,2,3,4$ can establish it for all $n$ . The right tool is the principle of mathematical induction: verify the smallest case, then show that whenever the statement holds for some value $k$ it is forced to hold for $k+1$ ; the two steps together certify it for every positive integer, since the chain of implications $1 \to 2 \to 3 \to \cdots$ can never break. The goal here is to carry out that argument for $P(n): (a+b)^n = \sum_{k=0}^{n} {}^{n}C_k\, a^{n-k}b^k$ , using only algebraic expansion and the Pascal identity relating consecutive rows of binomial coefficients.

Step 1. Check the base case $n=1$ . $P(1)$ reads $(a+b)^1 = {}^{1}C_0\,a^1 + {}^{1}C_1\,b^1$ . Since $^{1}C_0 = {}^{1}C_1 = 1$ , the right side is $a+b$ , matching the left side exactly. So $P(1)$ is true.

Step 2. Assume $P(k)$ true for some positive integer $k$ — the induction hypothesis, only assumed for the argument, not proved here: $(a+b)^k = {}^{k}C_0\,a^k + {}^{k}C_1\,a^{k-1}b + \cdots + {}^{k}C_{k-1}\,a\,b^{k-1} + {}^{k}C_k\,b^k$ . The task is to show this forces $P(k+1)$ true as well.

Step 3. Multiply both sides by $(a+b)$ , since $(a+b)^{k+1} = (a+b)(a+b)^k$ . Distributing across every term and collecting terms with the same power of $a$ and $b$ gives coefficients $\left({}^{k}C_r + {}^{k}C_{r-1}\right)$ multiplying $a^{k+1-r}b^r$ for each $r$ from $1$ to $k$ , plus the two untouched end terms $a^{k+1}$ and $b^{k+1}$ . This grouping is exactly how the addition rule of Pascal's triangle arises algebraically: two neighbouring entries of row $k$ sum to an entry of row $k+1$ .

Step 4. Apply the identity $^{k}C_r + {}^{k}C_{r-1} = {}^{k+1}C_r$ (the addition rule of Pascal's triangle) to every grouped coefficient from Step 3, together with $^{k+1}C_0 = 1 = {}^{k}C_0$ and $^{k+1}C_{k+1} = 1 = {}^{k}C_k$ . The result is exactly $(a+b)^{k+1} = {}^{k+1}C_0\,a^{k+1} + {}^{k+1}C_1\,a^k b + \cdots + {}^{k+1}C_{k+1}\,b^{k+1}$ , which is $P(k+1)$ . Since $P(1)$ holds and $P(k)$ true forces $P(k+1)$ true for every $k$ , the principle of mathematical induction guarantees $P(n)$ for every positive integer $n$ . Verified numerically: expanding $(x+2)^6$ with coefficients $^{6}C_0,\ldots,{}^{6}C_6 = 1,6,15,20,15,6,1$ gives $x^6+12x^5+60x^4+160x^3+240x^2+192x+64$ , matching direct repeated multiplication.

Go deeper · Derivation B · 1 minUsing the general term to solve for a hidden power: the standard method for a term independent of x or of a chosen power

Derivation BUsing the general term to solve for a hidden power: the standard method for a term independent of x or of a chosen power

Many competitive problems disguise one question in different clothing: "find the term independent of $x$ ," "find the coefficient of $x^5$ ," "find the term containing $x^{-3}$ " — all solved by the same procedure built on the general-term formula $T_{r+1} = {}^{n}C_r\, a^{n-r}b^r$ . The goal here is to derive that procedure for any binomial of the form $\left(\alpha x^{p} + \beta x^{-q}\right)^n$ with constants $\alpha, \beta$ and positive exponents $p, q$ , so it can be applied mechanically to any specific numbers a problem supplies.

Step 1. Write the general term for $\left(\alpha x^{p} + \beta x^{-q}\right)^n$ using formula $(3)$ with $a = \alpha x^p$ , $b = \beta x^{-q}$ : $T_{r+1} = {}^{n}C_r\,(\alpha x^{p})^{n-r}(\beta x^{-q})^{r} = {}^{n}C_r\,\alpha^{n-r}\beta^{r}\, x^{p(n-r) - qr}$ . The numerical part and the power of $x$ , $p(n-r)-qr = pn-(p+q)r$ , are now cleanly separated — this turns a question about the whole term into a question about a single exponent.

Step 2. Translate the target condition into an equation in $r$ alone. For the term independent of $x$ , set the exponent to zero: $pn - (p+q)r = 0$ , so $r = \dfrac{pn}{p+q}$ . For the coefficient of $x^m$ , set the exponent to $m$ instead: $r = \dfrac{pn-m}{p+q}$ . Either way the condition collapses to one linear equation in $r$ .

Step 3. Solve for $r$ and check admissibility. Since $r$ must be a whole number with $0 \le r \le n$ , any negative, fractional, or too-large solution means the requested term simply does not exist — a legitimate answer, not an error. A valid integer $r$ is substituted back into $^{n}C_r\,\alpha^{n-r}\beta^{r}$ to get the term's actual value.

Step 4. Read off the answer: $T_{r+1} = {}^{n}C_r\,\alpha^{n-r}\beta^r$ (independent-of- $x$ case) or the same expression times $x^m$ (coefficient-of- $x^m$ case). This chain — write the general term, isolate the exponent, solve for integer $r$ , substitute back — is applied verbatim in the Competitive-level worked example below, with $p=2, q=1, \alpha=\tfrac32, \beta=-\tfrac13, n=9$ . Verified: for $\left(\frac32 x^2 - \frac13 x^{-1}\right)^9$ , Step 2 gives $18-3r=0 \Rightarrow r=6$ , admissible, matching the worked example's answer $\tfrac{7}{18}$ .

Go deeper · Derivation C · 1 minWhy the middle term carries the greatest binomial coefficient

Derivation CWhy the middle term carries the greatest binomial coefficient

Formula $(4)$ asserted, without proof, that the "middle" term (or terms) is exactly where the largest binomial coefficient sits. That claim deserves a rigorous check rather than trust in the shape of Pascal's-triangle rows alone, because the same ratio-comparison technique reappears whenever a problem asks for the greatest term (not just coefficient) of a numerical binomial expansion. The goal is to compare each binomial coefficient to its neighbour and find exactly where the sequence $^{n}C_0, {}^{n}C_1, \ldots, {}^{n}C_n$ stops increasing and starts decreasing.

Step 1. Form the ratio of consecutive binomial coefficients. Using $^{n}C_r = \dfrac{n!}{r!(n-r)!}$ , simplification gives $\dfrac{{}^{n}C_{r+1}}{{}^{n}C_r} = \dfrac{n-r}{r+1}$ for $0 \le r \le n-1$ . This ratio controls whether the coefficients are rising or falling as $r$ increases by one.

Step 2. Decide when the ratio exceeds $1$ , since $^{n}C_{r+1} > {}^{n}C_r$ exactly then. Solving $\dfrac{n-r}{r+1} > 1$ gives $n-r > r+1$ , i.e. $r < \dfrac{n-1}{2}$ . So the coefficients strictly increase while $r < \frac{n-1}{2}$ and strictly decrease once $r$ exceeds it — a single rise, peak, and fall (a unimodal sequence).

Step 3. Locate the peak for each parity of $n$ . If $n=2m$ (even), $\frac{n-1}{2}=m-\frac12$ is a half-integer, so the unique peak is at $r=m=\frac{n}{2}$ , the term $T_{m+1}=T_{\frac{n}{2}+1}$ — the single middle term of $(4)$ . If $n=2m+1$ (odd), $\frac{n-1}{2}=m$ is a whole number, and substituting $r=m$ into the ratio gives $\dfrac{m+1}{m+1}=1$ exactly — the coefficients $^{n}C_m$ and $^{n}C_{m+1}$ are equal, and both exceed every neighbour.

Step 4. Translate the odd case to term positions: with $m=\frac{n-1}{2}$ , the tied maxima $^{n}C_m$ and $^{n}C_{m+1}$ correspond to $T_{m+1}=T_{\frac{n+1}{2}}$ and $T_{m+2}=T_{\frac{n+3}{2}}$ — exactly the two middle terms of $(4)$ . This confirms, by ratio argument rather than a picture, that the middle term (or tied pair) always carries the strictly largest binomial coefficient. Verified: for $n=8$ , the ratio test peaks at $r=4$ , matching $^{8}C_4=70$ , the tallest bar above; for $n=9$ , $m=4$ gives tied coefficients $^{9}C_4={}^{9}C_5=126$ , the two middle terms $T_5, T_6$ .

Worked examples

Board level - expanding a binomial with algebraic fractions (Example 1 pattern)

Expand $\left(x^2 + \dfrac{3}{x}\right)^4$ , $x \neq 0$ .

Step 1. Identify $a = x^2$ , $b = \dfrac{3}{x}$ and $n=4$ , and write out the theorem's five terms ( $n+1=5$ terms) using the row $1,4,6,4,1$ of Pascal's triangle for $^{4}C_0,\ldots,{}^{4}C_4$ : $(x^2+\tfrac3x)^4 = {}^{4}C_0(x^2)^4 + {}^{4}C_1(x^2)^3\!\left(\tfrac3x\right) + {}^{4}C_2(x^2)^2\!\left(\tfrac3x\right)^2 + {}^{4}C_3(x^2)\!\left(\tfrac3x\right)^3 + {}^{4}C_4\!\left(\tfrac3x\right)^4$ .

Step 2. Simplify each term one at a time, being careful with the exponent arithmetic on $x$ . First term: $1 \cdot x^8 = x^8$ . Second term: $4 \cdot x^6 \cdot \tfrac3x = 12x^5$ . Third term: $6 \cdot x^4 \cdot \tfrac{9}{x^2} = 54x^2$ .

Step 3. Continue with the last two terms. Fourth term: $4 \cdot x^2 \cdot \tfrac{27}{x^3} = \tfrac{108}{x}$ . Fifth term: $1 \cdot \tfrac{81}{x^4} = \tfrac{81}{x^4}$ . Every exponent of $x$ in the five terms — $8, 5, 2, -1, -4$ — decreases by exactly $3$ from one term to the next, an internal consistency check confirming no arithmetic slip.

Step 4. Add the five simplified terms in order. Answer: $\left(x^2+\dfrac{3}{x}\right)^4 = x^8 + 12x^5 + 54x^2 + \dfrac{108}{x} + \dfrac{81}{x^4}$ .

Board level - computing a large power using the binomial theorem (Example 2 pattern)

Compute $(98)^5$ using the Binomial Theorem.

Step 1. Rewrite $98$ as a sum or difference of two numbers whose own powers are effortless to compute: $98 = 100 - 2$ . So $(98)^5 = (100-2)^5$ , a binomial to the fifth power with $a=100$ , $b=2$ , and alternating signs since it is a difference.

Step 2. Expand using the row $1,5,10,10,5,1$ for $^{5}C_0,\ldots,{}^{5}C_5$ , with alternating signs from the $(x-y)^n$ special case: $(100-2)^5 = {}^{5}C_0(100)^5 - {}^{5}C_1(100)^4(2) + {}^{5}C_2(100)^3(2)^2 - {}^{5}C_3(100)^2(2)^3 + {}^{5}C_4(100)(2)^4 - {}^{5}C_5(2)^5$ .

Step 3. Evaluate each of the six terms: $10{,}000{,}000{,}000 - 5(100{,}000{,}000)(2) + 10(1{,}000{,}000)(4) - 10(10{,}000)(8) + 5(100)(16) - 32$ , which is $10{,}000{,}000{,}000 - 1{,}000{,}000{,}000 + 40{,}000{,}000 - 800{,}000 + 8{,}000 - 32$ .

Step 4. Add the positive terms and subtract the negative terms in order: the positives sum to $10{,}040{,}008{,}000$ and the negatives sum to $1{,}000{,}800{,}032$ , and $10{,}040{,}008{,}000 - 1{,}000{,}800{,}032 = 9{,}039{,}207{,}968$ . Answer: $(98)^5 = 9{,}039{,}207{,}968$ .

Competitive level (JEE / NEET) - term independent of x in a binomial expansion

Find the term independent of $x$ in the expansion of $\left(\dfrac{3}{2}x^2 - \dfrac{1}{3x}\right)^9$ .

Step 1. Set up the general term with $a = \tfrac32 x^2$ , $b = -\tfrac{1}{3x} = -\tfrac13 x^{-1}$ and $n=9$ : $T_{r+1} = {}^{9}C_r\left(\tfrac32 x^2\right)^{9-r}\left(-\tfrac13 x^{-1}\right)^{r} = {}^{9}C_r\left(\tfrac32\right)^{9-r}\left(-\tfrac13\right)^{r} x^{2(9-r)-r}$ .

Step 2. Isolate the exponent of $x$ : $2(9-r) - r = 18 - 3r$ . The term is independent of $x$ exactly when this exponent is zero, so $18 - 3r = 0 \Rightarrow r = 6$ , a whole number satisfying $0 \le 6 \le 9$ , so a genuine term of the expansion exists.

Step 3. Substitute $r=6$ into the numerical part: $T_7 = {}^{9}C_6\left(\tfrac32\right)^{3}\left(-\tfrac13\right)^{6} = 84 \times \tfrac{27}{8} \times \tfrac{1}{729}$ , using $^{9}C_6 = {}^{9}C_3 = 84$ and noting the even power $6$ keeps the sign positive despite the base being negative.

Step 4. Simplify the product carefully: $84 \times \dfrac{27}{8} \times \dfrac{1}{729} = \dfrac{84 \times 27}{8 \times 729} = \dfrac{2268}{5832} = \dfrac{7}{18}$ after cancelling the common factor of $324$ . Answer: the term independent of x is the 7th term, $T_7 = \dfrac{7}{18}$ . Timing note: budget about ninety seconds. The efficient order is exponent first: write only the power of $x$ from the general term, set it to zero, and solve for $r$ before any numerical arithmetic — plugging in numbers before finding $r$ wastes time on terms you will discard. Watch two traps: forgetting the negative sign inside $\left(-\tfrac13 x^{-1}\right)^r$ (it stays positive only because $r=6$ is even), and misreading $\dfrac{1}{3x}$ as $\dfrac{x}{3}$ , which silently reverses the exponent's sign and points you to the wrong value of $r$ .

Common traps

The most common error is the off-by-one slip in the general term: writing $T_r$ instead of $T_{r+1}$ for the term containing $b^r$ , which miscounts every subsequent step of a "find the $k$ th term" or "find the term independent of $x$ " problem. A related mistake is applying the even- $n$ middle-term rule (one middle term) to an odd $n$ , or vice versa — always check the parity of $n$ first. Sign errors plague the $(x-y)^n$ expansion: students forget the sign alternates term by term, or double-flip it when $y$ is already negative. In a binomial like $\left(\alpha x^p + \beta x^{-q}\right)^n$ , a frequent slip is computing the exponent of $x$ as $p(n-r)+qr$ instead of $p(n-r)-qr$ , forgetting a negative exponent subtracts, not adds. Students also confuse the binomial coefficient $^{n}C_r$ (the pure combinatorial number) with the numerical coefficient of a term, which also folds in any constants attached to $a$ and $b$ — the two agree only when $a$ and $b$ are bare variables. Finally, an inadmissible solved $r$ (negative, fractional, or larger than $n$ ) means no such term exists — not a sign of an arithmetic error to keep hunting for.

Common trapThe general term is T subscript r plus one, not T subscript r

The single biggest misconception in this chapter is misindexing the general term. The formula $T_{r+1} = {}^{n}C_r\, a^{n-r}b^r$ names the term at position $r+1$ in the expansion, counting from the first term as position $1$ . To find the fifth term, you must set $r+1=5$ , i.e. $r=4$ , and substitute $r=4$ into the formula — never set $r=5$ directly and never call the result " $T_r$ ." This single indexing slip, repeated across "find the middle term," "find the coefficient of $x^m$ ," and "find the term independent of $x$ " problems, is responsible for more lost marks in this chapter than every other error combined. Before submitting any answer that names a term by position, explicitly write the line " $r+1 = (\text{position asked for})$ " and solve for $r$ first.

You should be able to…

State the Binomial Theorem for a positive integral index $n$ , in both expanded and summation form, and explain why it has exactly $n+1$ terms.
Build Pascal's triangle row by row using simple addition, and rewrite any row using the binomial coefficients $^{n}C_r$ .
Prove the Binomial Theorem by mathematical induction, correctly using the identity $^{k}C_r + {}^{k}C_{r-1} = {}^{k+1}C_r$ to pass from $n=k$ to $n=k+1$ .
Write down the general term $T_{r+1} = {}^{n}C_r\,a^{n-r}b^r$ for any binomial expansion, and correctly solve for $r$ given a required term position.
Locate the middle term (even $n$ ) or the two middle terms (odd $n$ ) of an expansion, and explain why they carry the greatest binomial coefficient.
Expand $(x-y)^n$ and $(1+x)^n$ correctly, including the alternating signs of the first and the pure powers of $x$ in the second.
Derive and use the identities $2^n = \sum_r {}^{n}C_r$ and $0 = \sum_r (-1)^r\,{}^{n}C_r$ .
Compute large powers such as $(98)^5$ or $(101)^6$ exactly, by splitting the base into a convenient sum or difference and applying the theorem.
Find the term independent of a variable, or the coefficient of a specified power, in a binomial expansion involving fractional or negative exponents.
Use the Binomial Theorem to prove simple divisibility or remainder statements, such as showing $6^n - 5^n$ always leaves remainder $1$ on division by $25$ .

Ask the tutor: Want to work through locating a tricky specific term or the middle term of a harder binomial expansion, step by step together?

Why is the general term written as T r plus one and not simply T r?

Because the summation defining the Binomial Theorem starts its counting index $r$ at $0$ , not $1$ , while the terms themselves are naturally numbered starting from the first term as position $1$ . The term corresponding to $r=0$ is the very first term of the expansion, so it must be called $T_1$ , which forces the general rule $T_{r+1}$ rather than $T_r$ : the term carrying $b^r$ sits at position $r+1$ . Getting this shift right is essential the moment a problem asks for a specific numbered term — say the sixth term — because you must solve $r+1=6$ , giving $r=5$ , and only then substitute $r=5$ into $^{n}C_r\,a^{n-r}b^r$ . Treating the term index and the summation index as interchangeable is the most common error in this entire chapter.

How do I find the term independent of x without expanding the whole binomial?

Write down only the general term $T_{r+1} = {}^{n}C_r\,a^{n-r}b^r$ for the given binomial, and simplify just the power of $x$ that results, ignoring the numerical coefficient for the moment. Set that power of $x$ equal to zero, since "independent of $x$ " means the term has no $x$ left in it at all, and solve the resulting linear equation for $r$ . If $r$ comes out as a whole number between $0$ and $n$ inclusive, substitute it back into the full general term to get the actual constant value; if $r$ is not a valid integer in that range, the expansion simply has no term independent of $x$ , which is itself a complete and correct answer. This method, worked in full generality in the second derivation above, never requires writing out any other term of the expansion.

What is the difference between the binomial coefficient and the coefficient of a term?

The binomial coefficient is the pure combinatorial number $^{n}C_r$ that comes straight out of Pascal's triangle — it depends only on $n$ and $r$ , never on what $a$ and $b$ actually are. The coefficient of a term (sometimes called the numerical coefficient) is the entire numerical multiplier of the term after substituting the actual values or constants attached to $a$ and $b$ , which includes the binomial coefficient but also any extra constants riding along with $a$ and $b$ . For example, in $\left(\tfrac32 x^2 - \tfrac13 x^{-1}\right)^9$ , the binomial coefficient of the term with $r=6$ is $^{9}C_6 = 84$ , but the coefficient of that actual term (after multiplying in $\left(\tfrac32\right)^3$ and $\left(-\tfrac13\right)^6$ ) works out to $\tfrac{7}{18}$ — a very different number. Always read a question carefully to see which of the two it is asking for.

How do I know whether an expansion has one middle term or two?

Check the parity of the index $n$ , not the number of terms directly. If $n$ is even, the total number of terms $n+1$ is odd, so there is a single, uniquely defined middle term at position $\tfrac{n}{2}+1$ . If $n$ is odd, the total number of terms $n+1$ is even, so there is no single centre, and the two terms flanking the middle — at positions $\tfrac{n+1}{2}$ and $\tfrac{n+3}{2}$ — share the distinction, and (as proved in the third derivation above) they always carry exactly equal, and exactly the largest, binomial coefficients. A quick sanity check: $(a+b)^{10}$ has $11$ terms and one middle term $T_6$ ; $(a+b)^{9}$ has $10$ terms and two middle terms $T_5, T_6$ .

Can the Binomial Theorem be used for negative or fractional powers like (a plus b) to the power minus 2?

Not the version taught in this chapter. Everything proved here — the finite sum with $n+1$ terms, the binomial coefficients $^{n}C_r$ , the induction proof — depends essentially on $n$ being a positive integer, since the induction argument and the formula $^{n}C_r = \tfrac{n!}{r!(n-r)!}$ both require a whole, positive index. For negative or fractional indices the expansion becomes an infinite series with a different (though related) coefficient pattern, valid only when $|b/a| < 1$ ; that general Binomial Theorem is a topic for later study and is deliberately outside the scope of this NCERT chapter, which restricts itself to positive integral indices only.