General term and binomial coefficients

Expanding $(a+b)^{10}$ term by term to find just one coefficient is slow and error-prone. The general term formula skips straight to any single term of a binomial expansion — the 4th term, the term containing $x^5$ , or the middle term — without writing out the rest. It sits inside the Binomial Theorem chapter of Unit II (Algebra) in CBSE Class 11, built directly on the expansion $(a+b)^n = \sum {}^nC_k\,a^{n-k}b^k$ . The idea is compact but powerful: one formula, correctly indexed, answers "which term?", "what coefficient?" and "which term is in the middle?" questions that would otherwise need a full expansion. This article restates the formula, derives it from the binomial theorem, works a fresh middle-term example beyond the Pass-1 basics, and flags the indexing slip that costs the most marks.

The core insight

The $(r+1)$ -th term of $(a+b)^n$ is $T_{r+1} = {}^nC_r\,a^{n-r}b^r$ , so any specified term or coefficient can be picked out directly by choosing the right value of $r$ .

The core idea

The binomial theorem expands $(a+b)^n$ as a sum of $n+1$ terms, each built from a binomial coefficient and falling/rising powers of $a$ and $b$ :

(a+b)^n = \sum_{k=0}^{n} {}^nC_k\, a^{n-k}b^k

where: $n$ is the (positive integer) index of the expansion, and $k$ runs from $0$ to $n$ , giving one term per value of $k$ . Pulling out the term for a single value of $k = r$ gives the general term, conventionally labelled as the $(r+1)$ -th term of the expansion:

T_{r+1} = {}^nC_r\, a^{n-r}\,b^r

where: $r = 0, 1, 2, \dots, n$ ; $T_1$ is the first term ( $r=0$ ), $T_2$ is the second ( $r=1$ ), and so on — the subscript is always one more than $r$ .

Binomial coefficients. The numbers ${}^nC_r = \dfrac{n!}{r!(n-r)!}$ that multiply each term carry three properties worth knowing cold:

Symmetry: ${}^nC_r = {}^nC_{n-r}$ — the coefficients read the same from either end of the row.
End values: ${}^nC_0 = {}^nC_n = 1$ — the first and last terms are always just $a^n$ and $b^n$ .
Pascal's rule: ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$ — each coefficient is the sum of the two above it in the next row down, which is exactly how Pascal's triangle is built.
Row sum: ${}^nC_0 + {}^nC_1 + \cdots + {}^nC_n = 2^n$ , obtained by setting $a=b=1$ in the theorem.

Figure

Pascals triangle rows zero through five with each entry the sum of the two numbers above it

Why bother with the general term. To find the 6th term of $(2x-3)^9$ , substitute $r=5$ directly into $T_{r+1} = {}^9C_r(2x)^{9-r}(-3)^r$ — no need to expand all ten terms. To find a coefficient, set up the general term, solve for the value of $r$ that produces the required power of $x$ , then substitute back.

Middle term(s). This refinement builds directly on the general term and is formative-only this year, but it is worth understanding: when $n$ is even, $(a+b)^n$ has exactly one middle term, $T_{\frac{n}{2}+1}$ , since there are $n+1$ (odd) terms total. When $n$ is odd, there are two middle terms, $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$ , because $n+1$ is even and splits evenly around the centre.

Derivation ADeriving the general term from the binomial theorem

The goal is to show precisely how the single formula $T_{r+1} = {}^nC_r\,a^{n-r}b^r$ falls straight out of the summation form of the binomial theorem, so that the indexing convention — why it is called the $(r+1)$ -th term and not the $r$ -th — is understood rather than memorised.

Step 1. Start from the binomial theorem in summation form, $(a+b)^n = \sum_{k=0}^{n} {}^nC_k\,a^{n-k}b^k$ . This says the full expansion is built by letting $k$ run through every integer from $0$ to $n$ and adding up the corresponding term ${}^nC_k\,a^{n-k}b^k$ each time.

Step 2. Count how the terms line up with $k$ . When $k=0$ , the term is ${}^nC_0\,a^n b^0 = a^n$ , which is the first term of the expansion, $T_1$ . When $k=1$ , the term is ${}^nC_1\,a^{n-1}b^1$ , the second term, $T_2$ . In general, the term for a given value of $k$ occupies position $k+1$ in the listed expansion, because the count starts at $T_1$ (position 1) while $k$ starts at $0$ .

Step 3. Rename the running index $k$ as $r$ to match the term we want, and write the position explicitly: the term at $k=r$ is the $(r+1)$ -th term of the expansion, so it is labelled $T_{r+1}$ . Substituting $k=r$ into the summand from Step 1 gives $T_{r+1} = {}^nC_r\,a^{n-r}b^r$ , exactly the general term formula.

Step 4. Read off the indexing rule that follows immediately: to get the $p$ -th term of the expansion, set $r+1=p$ , i.e. use $r = p-1$ in the formula, not $r=p$ . This single substitution rule is the entire content of the "general term," and every worked problem below is just this formula with specific $a$ , $b$ , $n$ and $r$ plugged in.

Worked example

Find the middle term(s) in the expansion of $\left(x + \dfrac{1}{x}\right)^{8}$ , and also state the coefficient of $x^0$ in this expansion.

Step 1. Identify $n$ and the number of terms. Here $n=8$ is even, so the expansion has $n+1=9$ terms, and there is exactly one middle term, the 5th term, $T_{\frac{n}{2}+1} = T_5$ . In the general term formula this corresponds to $r = 4$ (since $T_5 = T_{r+1}$ needs $r=4$ ).

Step 2. Write the general term with $a=x$ , $b=\dfrac{1}{x}$ , $n=8$ : $T_{r+1} = {}^8C_r\,x^{8-r}\left(\dfrac{1}{x}\right)^r = {}^8C_r\,x^{8-2r}$ .

Step 3. Substitute $r=4$ to get the middle term: $T_5 = {}^8C_4\,x^{8-8} = {}^8C_4\,x^0 = {}^8C_4$ . Compute ${}^8C_4 = \dfrac{8!}{4!\,4!} = \dfrac{8\times7\times6\times5}{4\times3\times2\times1} = 70$ .

Step 4. Read off both answers together: since the middle term $T_5$ came out as $70\,x^0$ , it is simultaneously the term with zero power of $x$ . The middle term is $T_5 = 70$ , and the coefficient of $x^0$ in the expansion is also $70$ — here the two questions coincide because the exponent $8-2r$ vanishes exactly at the middle value $r=4$ .

Competitive stretch

If instead $n=9$ (odd) in $\left(x+\frac{1}{x}\right)^9$ , there are two middle terms, $T_5$ and $T_6$ , corresponding to $r=4$ and $r=5$ : $T_5 = {}^9C_4\,x^{9-2(4)} = {}^9C_4\,x^{1} = 126x$ , and $T_6 = {}^9C_5\,x^{9-2(5)} = {}^9C_5\,x^{-1} = \dfrac{126}{x}$ (using ${}^9C_4={}^9C_5=126$ by symmetry). Both middle terms come out with equal coefficients, as symmetry guarantees.

Common traps

Off-by-one indexing. For the $p$ -th term, use $r=p-1$ , not $r=p$ , in $T_{r+1}$ — this is the single most common slip in the whole topic.
Losing exponent signs with negative or fractional $b$ . Terms like $\left(\dfrac{1}{x}\right)^r = x^{-r}$ or $(-3)^r$ are easy to mishandle; write the exponent arithmetic out fully before simplifying.
Confusing a coefficient with the whole term. "The coefficient of $x^5$ " means the numerical factor alone, not the full term with $x^5$ attached.
Forcing an answer when no valid $r$ exists. Solving for the power of $x$ must give an integer $r$ with $0 \le r \le n$ ; if it doesn't, that power simply is not present in the expansion.

Ask the tutor: Not sure whether to use $r=p-1$ or $r=p$ for a specific term, or want another middle-term problem worked through? Ask the AI tutor to walk through one with you.

Why is the general term called T sub r plus 1 and not T sub r

Because the summation in the binomial theorem starts its counting index $k$ (or $r$ ) at $0$ , while the terms of the expansion are conventionally numbered starting at $1$ . The term corresponding to $r=0$ is the first term $T_1$ , so a shift of one is built in: the term at index $r$ is always the $(r+1)$ -th term. Keeping this offset straight is exactly what the derivation in this article works through step by step, and it is the source of almost every indexing mistake students make.

How do I quickly tell which r gives a specific power of x without long algebra

Write the general term for your specific expansion first, simplify the power of $x$ into a single expression in $r$ (as in $x^{8-2r}$ or $x^{20-3r}$ in the worked examples), then set that expression equal to the target power and solve the resulting linear equation for $r$ . If $r$ comes out as a non-integer or outside $0$ to $n$ , the power you want simply is not present — that is a valid, expected outcome, not a mistake to fix.

Do I need to memorise Pascal's triangle or can I always compute nCr directly

Either works, and both are worth having. Pascal's triangle is fast for small $n$ (up to about $n=8$ or so) and builds intuition for the symmetry and Pascal's-rule properties, since each row is generated by adding adjacent entries from the row above. For larger $n$ , or when you only need one or two specific coefficients rather than the whole row, computing ${}^nC_r = \dfrac{n!}{r!(n-r)!}$ directly is quicker and less error-prone than building out several rows of the triangle just to reach one number.

General term and binomial coefficients

The core idea

Worked example

Competitive stretch

Common traps

Before this