Binomial theorem

Multiply out $(x+y)^2$ by hand and you get $x^2+2xy+y^2$ in seconds. Try $(x+y)^{10}$ the same way — term by term, bracket by bracket — and you will be at it for an hour, and still risk a sign error near the end. The binomial theorem removes the grinding altogether: it hands you every coefficient in the expansion of $(x+y)^n$ directly, built from combinations $^nC_r$ you already met in Permutations and Combinations. The theorem does three things at once. It tells you exactly how many terms an expansion has, it gives a formula for any single term without expanding the rest, and it explains why the coefficients form the neat, symmetric rows of Pascal's triangle. This article proves the theorem by induction, shows how to pull out one specific term or coefficient on demand, and works two full examples — expanding a negative binomial and hunting a coefficient buried inside a fractional power — the two question types that dominate ISC Class 11 exams.

The core insight

$(x+y)^n$ expands as $\sum_{k=0}^n {}^nC_k x^{n-k}y^k$ , so every coefficient and every individual term is a combination $^nC_k$ that can be read off or picked out without expanding the whole bracket.

The core idea

The expansion. For a positive integer $n$ and real numbers $x,y$ :

(x+y)^n = \sum_{k=0}^{n} {}^nC_k\, x^{n-k} y^{k} = {}^nC_0 x^n + {}^nC_1 x^{n-1}y + \cdots + {}^nC_n y^n

where: $^nC_k = \dfrac{n!}{k!(n-k)!}$ is the number of ways to choose $k$ objects from $n$ , and it also happens to be the coefficient of the term carrying $y^k$ . The sum runs from $k=0$ to $k=n$ , so the expansion always has exactly $n+1$ terms, one more than the index.

The general term. Rather than writing out the whole sum, most problems only need one term. Writing the terms as $T_1, T_2, T_3,\ldots$ in order, the $(r+1)$ -th term is

T_{r+1} = {}^nC_r\, x^{n-r} y^{r}

This single formula is the workhorse of the whole chapter: set $x^{n-r}y^r$ equal to whatever power you are chasing, solve for $r$ , then substitute back to get the coefficient.

Middle term(s). If $n$ is even, the expansion has an odd number of terms ( $n+1$ ), so there is exactly one middle term, $T_{(n/2)+1}$ . If $n$ is odd, $n+1$ is even, so two terms sit in the middle, $T_{(n+1)/2}$ and $T_{(n+3)/2}$ . Checking the parity of $n$ first avoids applying the wrong rule.

Pascal's triangle. The coefficients $^nC_0,\ldots,{}^nC_n$ for successive $n$ build the familiar triangle, each entry the sum of the two above it: $^{n+1}C_r = {}^nC_{r-1} + {}^nC_r$ . For $n \le 6$ this is faster than computing factorials by hand.

Why the general term matters. Most exam questions do not ask for a full expansion — they ask for one specific term, one coefficient, or the term free of $x$ . Writing out all $n+1$ terms just to isolate one of them wastes time and invites arithmetic slips. The general term formula lets you jump straight to the power you want: match the exponent of the variable you are chasing to $n-r$ or $r$ in $T_{r+1}$ , solve the resulting linear equation for $r$ , and substitute that single value back in. This "solve for $r$ , then substitute" pattern is worth memorising as a routine, since it reappears in every "find the coefficient of…" or "find the term independent of $x$ " question.

Figure

Pascal triangle rows zero through five with each entry equal to the sum of the two entries diagonally above it

Proof by mathematical induction

The goal is to prove that $(x+y)^n = \sum_{k=0}^n {}^nC_k x^{n-k}y^k$ holds for every positive integer $n$ , using only the principle of mathematical induction and one identity from Pascal's triangle, $^mC_{r-1} + {}^mC_r = {}^{m+1}C_r$ , with no reference to calculus or any result beyond what is already available at this stage of the algebra syllabus.

Step 1. Verify the base case $n=1$ . The formula gives $(x+y)^1 = {}^1C_0 x + {}^1C_1 y = x+y$ , since $^1C_0={}^1C_1=1$ . This matches $(x+y)^1$ exactly, so the statement is true for $n=1$ .

Step 2. State the inductive hypothesis. Assume the formula holds for some positive integer $n=m$ , that is, assume $(x+y)^m = \sum_{k=0}^m {}^mC_k x^{m-k}y^k$ is true. This assumption is not proved separately; it is the standing supposition the induction step is built on.

Step 3. Multiply both sides by $(x+y)$ to reach $n=m+1$ . On the left, $(x+y)^m(x+y) = (x+y)^{m+1}$ . On the right, distribute $(x+y)$ across every term of the sum: $(x+y)\sum_{k=0}^m {}^mC_k x^{m-k}y^k = \sum_{k=0}^m {}^mC_k x^{m+1-k}y^k + \sum_{k=0}^m {}^mC_k x^{m-k}y^{k+1}$ .

Step 4. Collect the coefficient of $x^{m+1-r}y^r$ from the two sums. It picks up $^mC_r$ from the first sum (matching $k=r$ ) and $^mC_{r-1}$ from the second sum (matching $k=r-1$ , since that term carries $y^{k+1}=y^r$ ). So the combined coefficient is $^mC_{r-1} + {}^mC_r$ , which by Pascal's identity equals $^{m+1}C_r$ exactly.

Step 5. Conclude. Since every coefficient of $(x+y)^{m+1}$ collapses to $^{m+1}C_r$ , the expansion is $(x+y)^{m+1} = \sum_{r=0}^{m+1} {}^{m+1}C_r\, x^{m+1-r}y^r$ — the same formula with $n$ replaced by $m+1$ . The statement is true for $n=1$ and, whenever it is true for $n=m$ , it is true for $n=m+1$ ; by the principle of mathematical induction it is therefore true for every positive integer $n$ .

Worked example

Expand $(2x-3y)^4$ using the binomial theorem, then verify the result by substitution.

Step 1. Identify the pieces: treat $a=2x$ , $b=-3y$ , $n=4$ , and read the coefficients $1,4,6,4,1$ straight off row $n=4$ of Pascal's triangle. So $(a+b)^4 = {}^4C_0a^4+{}^4C_1a^3b+{}^4C_2a^2b^2+{}^4C_3ab^3+{}^4C_4b^4$ .

Step 2. Substitute $a=2x$ , $b=-3y$ into every term, keeping the sign of $b$ attached throughout: $(2x-3y)^4 = (2x)^4 + 4(2x)^3(-3y) + 6(2x)^2(-3y)^2 + 4(2x)(-3y)^3 + (-3y)^4$ .

Step 3. Simplify each term in turn: $(2x)^4=16x^4$ ; $4(2x)^3(-3y)=4\cdot8x^3\cdot(-3y)=-96x^3y$ ; $6(2x)^2(-3y)^2=6\cdot4x^2\cdot9y^2=216x^2y^2$ ; $4(2x)(-3y)^3=4\cdot2x\cdot(-27y^3)=-216xy^3$ ; $(-3y)^4=81y^4$ .

Step 4. Add the terms and verify. Summing gives $(2x-3y)^4 = 16x^4-96x^3y+216x^2y^2-216xy^3+81y^4$ . Substitute $x=y=1$ : the right side gives $16-96+216-216+81=1$ , and the left side gives $(2-3)^4=(-1)^4=1$ — the two agree, confirming the expansion.

Final answer: $(2x-3y)^4 = 16x^4-96x^3y+216x^2y^2-216xy^3+81y^4$ .

Competitive stretch

Find the coefficient of $x^7$ in $\left(x^2-\dfrac{2}{x}\right)^{11}$ . With $a=x^2$ , $b=-2/x$ , $n=11$ : $T_{r+1} = {}^{11}C_r(x^2)^{11-r}\left(-\dfrac{2}{x}\right)^r = {}^{11}C_r(-2)^r x^{22-3r}$ . Setting $22-3r=7$ gives $r=5$ , so $T_6 = {}^{11}C_5(-2)^5x^7$ . Since $^{11}C_5=462$ and $(-2)^5=-32$ , the coefficient is $462\times(-32) = \mathbf{-14784}$ .

Common traps

Losing track of signs. When the binomial has a negative term, always substitute $b$ with its full sign into $T_{r+1}$ before simplifying — odd powers of $b$ flip the sign, even powers do not.
Off-by-one in the term number. The term containing $y^r$ is $T_{r+1}$ , not $T_r$ ; asking for "the 5th term" means $r=4$ , a common trap when finding a specified term.
Mixing up the middle-term rule. Even $n$ gives one middle term; odd $n$ gives two. Check the parity of $n$ before applying either rule.
Forgetting $^nC_0={}^nC_n=1$ and mis-simplifying factorials; when $r$ is close to $n$ , compute $^nC_r$ as $^nC_{n-r}$ instead — it is usually faster.

Ask the tutor: Stuck on pulling out a specific term or coefficient, or want the induction proof walked through one step at a time — ask the AI tutor to work a binomial theorem problem with you.

How do I know whether to use nCr or nCr with the sign already built in when the binomial has a minus sign

Always keep the minus sign as part of $b$ itself, not as a separate factor multiplying $^nC_r$ . If the binomial is $(a-c)^n$ , write $b=-c$ and substitute the whole signed quantity into $T_{r+1}={}^nC_r a^{n-r}b^r$ ; the theorem's coefficient $^nC_r$ is always positive, and every sign in the final term comes purely from raising the negative $b$ to the power $r$ . This is why odd-power terms in an alternating expansion come out negative and even-power terms come out positive.

Why does putting x equals y equals 1 give 2 to the power n, and is that just a curiosity

Substituting $x=y=1$ into $(x+y)^n=\sum{}^nC_k x^{n-k}y^k$ turns every term into just $^nC_k$ , since any number raised to any power is still $1$ . So the sum collapses to $^nC_0+{}^nC_1+\cdots+{}^nC_n = 2^n$ . It is not just a curiosity — it is a genuinely useful check: after expanding a binomial, plugging in $x=y=1$ lets you verify that the sum of all coefficients matches $2^n$ , catching arithmetic slips before you submit an answer.

When a problem asks for the coefficient of a specific power like x to the seventh, do I need to expand the whole binomial

No, and doing so wastes time. Write down the general term $T_{r+1}={}^nC_r\,a^{n-r}b^r$ for the given binomial, simplify the power of the variable in terms of $r$ , set that power equal to the target exponent, and solve for $r$ . Substitute that single value of $r$ back into $T_{r+1}$ to get the one term you need, along with its coefficient. This general-term shortcut is exactly what the theorem is designed for — you never expand terms you do not need.