Binomial Theorem • Topic 2 of 3

General Term and Middle Term

Writing out a whole expansion is wasteful when a question asks for just one term. The general term is a single formula that produces any term you want by plugging in its position.

$$T_{r+1}=\binom{n}{r}a^{n-r}b^{r}$$

Here $T_{r+1}$ is the $(r+1)$th term of $(a+b)^n$. The off-by-one is the only thing to watch: $r=0$ gives the first term $T_1=a^n$, $r=1$ gives the second term, and so on up to $r=n$ for the last term $T_{n+1}=b^n$. So to reach the $k$th term, set $r=k-1$. For the common form $(1+x)^n$ the general term simplifies to $T_{r+1}=\binom{n}{r}x^{r}$.

How to use it. To find a particular term (say the $7$th), set $r=k-1=6$, substitute, and simplify the numbers and powers. To find the coefficient of a given power $x^m$, first collect every power of $x$ in the general term into one exponent, set that exponent equal to $m$, solve the resulting linear equation for $r$, and only then evaluate $\binom{n}{r}$ together with any constants. To find the term independent of $x$ (the constant term) you do the same but set the combined exponent equal to $0$.

Counting terms from the end. Since there are $n+1$ terms, the $k$th term from the end is the $(n-k+2)$th term from the start. Equivalently, the $k$th term from the end of $(a+b)^n$ is the $k$th term from the start of $(b+a)^n$ — a useful re-labelling that saves arithmetic.

The middle term depends on whether $n$ is even or odd, because there are $n+1$ terms in all:

$$\text{if }n\text{ is even: one middle term }T_{\frac{n}{2}+1};\qquad\text{if }n\text{ is odd: two middle terms }T_{\frac{n+1}{2}}\text{ and }T_{\frac{n+3}{2}}.$$

The logic is simple. With $n+1$ terms total, an even $n$ gives an odd count of terms, so there is one term dead in the centre at position $\frac{n}{2}+1$. An odd $n$ gives an even count, leaving two terms sharing the middle. The middle term also carries the greatest binomial coefficient, which is why it matters in later work.

$n$Number of terms $(n+1)$Parity of $n$Middle term(s)
$4$$5$even$T_3$ (one)
$6$$7$even$T_4$ (one)
$5$$6$odd$T_3,\,T_4$ (two)
$7$$8$odd$T_4,\,T_5$ (two)

Deeper Insight — read $r$ as a counter, not a position: Almost every error with the general term comes from confusing the index $r$ with the term number. Think of $r$ as a counter that starts at zero — it counts how many factors of $b$ a term carries, and equivalently how many of the $n$ brackets contributed a $b$. That is exactly why $\binom{n}{r}$ appears: it counts the choices of those $r$ brackets, the same subset-counting from the previous topic. Because $r$ starts at $0$, the term number is always one more, hence $T_{r+1}$. Once you fix this in your mind, middle terms stop being a memorised rule: the centre of $n+1$ slots is at slot $\frac{n+1}{2}+\frac12=\frac{n}{2}+1$ when that is a whole number (even $n$), and otherwise the two slots straddling the centre are the middle pair. Treat the formula as a way of selecting a term and the parity question answers itself.

Anatomy of the general term T r plus one Anatomy of the General Term Tr+1 = C(n, r) an−r br binomial coefficientpower of a falls to n−rr counts the factors of b Middle term for even n versus odd n Where the Middle Sits EVEN n = 4 → 5 termsT₁T₂T₃T₄T₅one middle term ODD n = 5 → 6 termsT₁T₂T₃T₄T₅T₆two middle terms
Solved Examples
1
Worked Example
Find the 5th term in the expansion of $(2x+3)^7$.
Solution
  1. For the 5th term set $r=k-1=4$ in $T_{r+1}=\binom{n}{r}a^{n-r}b^{r}$ with $n=7,\ a=2x,\ b=3$.
  2. $T_5=\binom{7}{4}(2x)^{7-4}(3)^4=\binom{7}{4}(2x)^3(3)^4$.
  3. $\binom{7}{4}=35,\ (2x)^3=8x^3,\ 3^4=81$.
  4. $T_5=35\cdot 8x^3\cdot 81=22680\,x^3$.

Answer: $T_5=22680\,x^3$.

2
Worked Example
Find the middle term in the expansion of $\left(x+\dfrac{1}{x}\right)^{10}$.
Solution
  1. $n=10$ is even, so there is one middle term at position $\frac{n}{2}+1=6$, i.e. $r=5$.
  2. $T_6=\binom{10}{5}x^{10-5}\left(\dfrac{1}{x}\right)^{5}=\binom{10}{5}x^{5}\cdot x^{-5}$.
  3. $\binom{10}{5}=252$ and the powers of $x$ cancel.

Answer: The middle term is $T_6=252$.

3
Worked Example
Find both middle terms in the expansion of $(2x-y)^7$.
Solution
  1. $n=7$ is odd, so the two middle terms are $T_{\frac{n+1}{2}}=T_4$ and $T_{\frac{n+3}{2}}=T_5$.
  2. $T_4$ (here $r=3$): $\binom{7}{3}(2x)^4(-y)^3=35\cdot 16x^4\cdot(-y^3)=-560\,x^4y^3$.
  3. $T_5$ (here $r=4$): $\binom{7}{4}(2x)^3(-y)^4=35\cdot 8x^3\cdot y^4=280\,x^3y^4$.

Answer: Middle terms are $T_4=-560\,x^4y^3$ and $T_5=280\,x^3y^4$.

4
Worked Example
Find the coefficient of $x^5$ in the expansion of $(1+x)^{15}$.
Solution
  1. For $(1+x)^n$ the general term is $T_{r+1}=\binom{n}{r}x^{r}$, so the power of $x$ equals $r$.
  2. Set $r=5$ with $n=15$: coefficient $=\binom{15}{5}$.
  3. $\binom{15}{5}=\dfrac{15\cdot14\cdot13\cdot12\cdot11}{5!}=\dfrac{360360}{120}=3003$.

Answer: The coefficient of $x^5$ is $3003$.

5
Worked Example
Find the 6th term from the end in the expansion of $(a+b)^{12}$.
Solution
  1. There are $n+1=13$ terms; the 6th term from the end is the $(13-6+1)=8$th term from the start.
  2. For the 8th term set $r=7$: $T_8=\binom{12}{7}a^{12-7}b^{7}=\binom{12}{7}a^5b^7$.
  3. $\binom{12}{7}=\binom{12}{5}=792$.

Answer: The 6th term from the end is $T_8=792\,a^5b^7$.

6
Worked Example
Find the middle term in the expansion of $\left(\dfrac{x}{3}+9y\right)^{8}$.
Solution
  1. $n=8$ is even, so the single middle term is at $\frac{n}{2}+1=5$, i.e. $r=4$.
  2. $T_5=\binom{8}{4}\left(\dfrac{x}{3}\right)^{4}(9y)^{4}=70\cdot\dfrac{x^4}{81}\cdot 6561\,y^4$.
  3. $\dfrac{6561}{81}=81$, so $T_5=70\cdot 81\,x^4y^4=5670\,x^4y^4$.

Answer: The middle term is $T_5=5670\,x^4y^4$.

7
Worked Example
Find the coefficient of $x^7$ in the expansion of $\left(x^2+\dfrac{1}{x}\right)^{11}$.
Solution
  1. $T_{r+1}=\binom{11}{r}(x^2)^{11-r}\left(\dfrac{1}{x}\right)^{r}=\binom{11}{r}x^{22-2r-r}=\binom{11}{r}x^{22-3r}$.
  2. Set $22-3r=7\Rightarrow 3r=15\Rightarrow r=5$.
  3. Coefficient $=\binom{11}{5}=462$.

Answer: The coefficient of $x^7$ is $462$.

8
Worked Example
Find the term independent of $x$ in the expansion of $\left(x-\dfrac{2}{x^2}\right)^{9}$.
Solution
  1. $T_{r+1}=\binom{9}{r}x^{9-r}\left(-\dfrac{2}{x^2}\right)^{r}=\binom{9}{r}(-2)^{r}x^{9-r-2r}=\binom{9}{r}(-2)^{r}x^{9-3r}$.
  2. For the constant term set $9-3r=0\Rightarrow r=3$.
  3. Term $=\binom{9}{3}(-2)^{3}=84\cdot(-8)=-672$.

Answer: The term independent of $x$ is $-672$.

9
Worked Example
Find the coefficient of $x^4$ in the expansion of $(1+2x)^6$.
Solution
  1. $T_{r+1}=\binom{6}{r}(2x)^{r}=\binom{6}{r}2^{r}x^{r}$, so the power of $x$ is $r$.
  2. Set $r=4$: coefficient $=\binom{6}{4}2^{4}$.
  3. $=15\cdot 16=240$.

Answer: The coefficient of $x^4$ is $240$.

10
Worked Example
Find the 4th term in the expansion of $\left(\dfrac{a}{b}-\dfrac{b}{a}\right)^{8}$.
Solution
  1. For the 4th term set $r=3$ with $a$-part $\dfrac{a}{b}$ and $b$-part $-\dfrac{b}{a}$.
  2. $T_4=\binom{8}{3}\left(\dfrac{a}{b}\right)^{5}\left(-\dfrac{b}{a}\right)^{3}=56\cdot\dfrac{a^5}{b^5}\cdot\left(-\dfrac{b^3}{a^3}\right)$.
  3. $=-56\cdot\dfrac{a^{5-3}}{b^{5-3}}=-56\,\dfrac{a^2}{b^2}$.

Answer: $T_4=-56\,\dfrac{a^2}{b^2}$.

11
Worked Example
Find the middle term(s) in the expansion of $\left(3-\dfrac{x^3}{6}\right)^{7}$.
Solution
  1. $n=7$ is odd, so there are two middle terms: $T_4$ ($r=3$) and $T_5$ ($r=4$), with $a=3,\ b=-\dfrac{x^3}{6}$.
  2. $T_4=\binom{7}{3}3^{4}\left(-\dfrac{x^3}{6}\right)^{3}=35\cdot 81\cdot\left(-\dfrac{x^9}{216}\right)=-\dfrac{2835}{216}x^9=-\dfrac{105}{8}x^9$.
  3. $T_5=\binom{7}{4}3^{3}\left(-\dfrac{x^3}{6}\right)^{4}=35\cdot 27\cdot\dfrac{x^{12}}{1296}=\dfrac{945}{1296}x^{12}=\dfrac{35}{48}x^{12}$.

Answer: Middle terms: $T_4=-\dfrac{105}{8}x^9$ and $T_5=\dfrac{35}{48}x^{12}$.

12
Worked Example
If the coefficients of the $5$th and $6$th terms in $(1+x)^n$ are equal, find $n$.
Solution
  1. The $5$th term has $r=4$ (coefficient $\binom{n}{4}$); the $6$th has $r=5$ (coefficient $\binom{n}{5}$).
  2. Set them equal: $\binom{n}{4}=\binom{n}{5}$. Two binomial coefficients are equal when their lower indices add to $n$: $4+5=n$.
  3. So $n=9$ (check: $\binom{9}{4}=\binom{9}{5}=126$).

Answer: $n=9$.

Key Points

  • The general term is $T_{r+1}=\binom{n}{r}a^{n-r}b^{r}$; to get the $k$th term set $r=k-1$.
  • For $(1+x)^n$ the general term reduces to $T_{r+1}=\binom{n}{r}x^{r}$, so the power of $x$ equals $r$.
  • To find the coefficient of $x^m$, collect the power of $x$ in $T_{r+1}$, set it equal to $m$, solve for $r$, then evaluate $\binom{n}{r}$ with any constants.
  • If $n$ is even there is one middle term $T_{\frac{n}{2}+1}$; if $n$ is odd there are two, $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$ — and the middle carries the greatest coefficient.
  • The $k$th term from the end is the $(n-k+2)$th term from the beginning.
  • $\binom{n}{r}=\binom{n}{s}$ forces either $r=s$ or $r+s=n$ — handy when two coefficients are set equal.
  • Read $r$ as a counter starting at $0$ — it records how many factors of $b$ the term carries.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The general term of $(a+b)^n$ is:
Explanation: $T_{r+1}={}^{n}C_{r}a^{n-r}b^{r}$.
Q2.The middle term of $(x+y)^8$ is the:
Explanation: For even $n=8$, middle is the $(n/2+1)=5$th term.
Q3.For odd $n$, the number of middle terms is:
Explanation: Two middle terms when $n$ is odd.
Q4.In $(1+x)^n$, the coefficient of $x^r$ is:
Explanation: $T_{r+1}={}^{n}C_{r}x^{r}$.
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