General Term and Middle Term in Binomial Expansion

Master the general term formula, middle term, and term finding techniques for JEE

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General Term and Middle Term

General Term Formula

The general term (also called the $(r+1)^{\text{th}}$ term) in the expansion of $(a + b)^n$ is:

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

where $r = 0, 1, 2, \ldots, n$

Key Points

  • $T_{r+1}$ represents the $(r+1)^{\text{th}}$ term
  • $r$ is the index starting from 0
  • First term: $T_1$ when $r = 0$
  • Last term: $T_{n+1}$ when $r = n$

General Term in Different Forms

Form 1: $(x + a)^n$

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

Form 2: $(1 + x)^n$

$$\boxed{T_{r+1} = \binom{n}{r} x^r}$$

Form 3: $(ax + b)^n$

$$\boxed{T_{r+1} = \binom{n}{r} (ax)^{n-r} b^r = \binom{n}{r} a^{n-r} b^r x^{n-r}}$$

Form 4: $\left(x + \frac{1}{x}\right)^n$

$$\boxed{T_{r+1} = \binom{n}{r} x^{n-r} \cdot x^{-r} = \binom{n}{r} x^{n-2r}}$$

Middle Term(s)

The middle term(s) depend on whether $n$ is even or odd.

Case 1: When $n$ is Even

Single Middle Term: $T_{\frac{n}{2} + 1}$ (the $\left(\frac{n}{2} + 1\right)^{\text{th}}$ term)

$$\boxed{T_{\frac{n}{2}+1} = \binom{n}{n/2} a^{n/2} b^{n/2}}$$

Example: In $(a + b)^6$, middle term is $T_4$ (when $r = 3$)

Interactive Demo: Visualize Binomial Coefficients

See how coefficients relate to Pascal’s Triangle positions.

Case 2: When $n$ is Odd

Two Middle Terms: $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$

$$\boxed{T_{\frac{n+1}{2}} = \binom{n}{(n-1)/2} a^{(n+1)/2} b^{(n-1)/2}}$$ $$\boxed{T_{\frac{n+3}{2}} = \binom{n}{(n+1)/2} a^{(n-1)/2} b^{(n+1)/2}}$$

Example: In $(a + b)^7$, middle terms are $T_4$ and $T_5$ (when $r = 3$ and $r = 4$)

Memory Tricks

🎯 General Term Index

“r+1 Rule”: The $(r+1)^{\text{th}}$ term has:

  • Binomial coefficient: $\binom{n}{r}$
  • First quantity’s power: $n - r$
  • Second quantity’s power: $r$

Mnemonic: “Name Remains, R Goes Up”

  • $n$ stays in $\binom{n}{r}$
  • $r$ increases from 0 to $n$
  • Powers of second term go up

🎯 Middle Term Quick Find

Even $n$: Middle position = $\frac{n}{2} + 1$

  • For $n = 6$: Position = $3 + 1 = 4$ (4th term)

Odd $n$: Two middles at $\frac{n+1}{2}$ and $\frac{n+1}{2} + 1$

  • For $n = 7$: Positions = $4$ and $5$ (4th and 5th terms)

🎯 Independent Term Finder

“Power Sum = 0”: For term independent of $x$ in $\left(x^a + \frac{b}{x^c}\right)^n$:

  • General term power: $a(n-r) - cr$
  • Set equal to 0: $a(n-r) - cr = 0$
  • Solve for $r$

Common Mistakes to Avoid

❌ Mistake 1: Confusing $r$ and Term Number

Wrong: “5th term means $r = 5$” ✗

Correct: “5th term means $T_5$, so $r = 4$” ✓

Remember: Term number = $r + 1$

❌ Mistake 2: Wrong Middle Term for Odd $n$

Wrong: For $(x + y)^9$, middle term is $T_5$ ✗

Correct: For $(x + y)^9$, there are TWO middle terms: $T_5$ and $T_6$ ✓

❌ Mistake 3: Power Calculation Error

Wrong: In $\left(2x^2 - \frac{3}{x}\right)^{10}$, power of $x$ in $T_{r+1}$ is $2r - 1$ ✗

Correct: $T_{r+1} = \binom{10}{r}(2x^2)^{10-r}\left(-\frac{3}{x}\right)^r$

Power of $x = 2(10-r) - r = 20 - 3r$ ✓

❌ Mistake 4: Sign Error in Negative Terms

Wrong: In $(x - 2y)^5$, $T_3 = \binom{5}{2}x^3(2y)^2$ ✗

Correct: $T_3 = \binom{5}{2}x^3(-2y)^2 = 10x^3 \cdot 4y^2 = 40x^3y^2$ ✓

Remember: Include the sign when raising to power!

Solved Examples

Example 1: Finding Specific Term (JEE Main)

Find the 7th term in the expansion of $(3x - 2y)^{10}$.

Solution: For 7th term: $T_7 = T_{6+1}$, so $r = 6$

$$\begin{align} T_7 &= \binom{10}{6}(3x)^{10-6}(-2y)^6 \\ &= \binom{10}{6}(3x)^4(64y^6) \\ &= 210 \times 81x^4 \times 64y^6 \\ &= 1088640x^4y^6 \end{align}$$

Example 2: Middle Term (JEE Main)

Find the middle term in the expansion of $(2x - 3y)^8$.

Solution: Since $n = 8$ (even), there is ONE middle term.

Middle term = $T_{\frac{8}{2}+1} = T_5$ (when $r = 4$)

$$\begin{align} T_5 &= \binom{8}{4}(2x)^{8-4}(-3y)^4 \\ &= 70 \times 16x^4 \times 81y^4 \\ &= 90720x^4y^4 \end{align}$$

Example 3: Term Independent of $x$ (JEE Advanced)

Find the term independent of $x$ in $\left(3x^2 - \frac{2}{x^3}\right)^{10}$.

Solution: General term:

$$T_{r+1} = \binom{10}{r}(3x^2)^{10-r}\left(-\frac{2}{x^3}\right)^r$$ $$= \binom{10}{r} \cdot 3^{10-r} \cdot (-2)^r \cdot x^{2(10-r)-3r}$$ $$= \binom{10}{r} \cdot 3^{10-r} \cdot (-2)^r \cdot x^{20-5r}$$

For term independent of $x$: $20 - 5r = 0$

$\Rightarrow r = 4$

$$\begin{align} T_5 &= \binom{10}{4} \cdot 3^6 \cdot (-2)^4 \\ &= 210 \times 729 \times 16 \\ &= 2449440 \end{align}$$

Example 4: Coefficient of Specific Power (JEE Advanced)

Find the coefficient of $x^{-5}$ in the expansion of $\left(x - \frac{1}{x^2}\right)^{11}$.

Solution: General term:

$$T_{r+1} = \binom{11}{r}(x)^{11-r}\left(-\frac{1}{x^2}\right)^r$$ $$= \binom{11}{r}(-1)^r x^{11-r-2r} = \binom{11}{r}(-1)^r x^{11-3r}$$

For $x^{-5}$: $11 - 3r = -5$

$\Rightarrow 3r = 16$

$\Rightarrow r = \frac{16}{3}$ (not an integer!)

Therefore, there is NO term containing $x^{-5}$. Coefficient = 0.

Example 5: Middle Terms in Odd Power (JEE Main)

Find both middle terms in the expansion of $\left(x + \frac{2}{x}\right)^7$.

Solution: Since $n = 7$ (odd), there are TWO middle terms.

First middle term: $T_{\frac{7+1}{2}} = T_4$ (when $r = 3$)

$$\begin{align} T_4 &= \binom{7}{3}(x)^{7-3}\left(\frac{2}{x}\right)^3 \\ &= 35 \times x^4 \times \frac{8}{x^3} \\ &= 280x \end{align}$$

Second middle term: $T_{\frac{7+3}{2}} = T_5$ (when $r = 4$)

$$\begin{align} T_5 &= \binom{7}{4}(x)^{7-4}\left(\frac{2}{x}\right)^4 \\ &= 35 \times x^3 \times \frac{16}{x^4} \\ &= \frac{560}{x} \end{align}$$

Practice Problems

Level 1: JEE Main Basics

Problem 1.1: Find the 5th term in $(2a + b)^8$.

Solution

$T_5 = T_{4+1}$, so $r = 4$

$$T_5 = \binom{8}{4}(2a)^4(b)^4 = 70 \times 16a^4 \times b^4 = 1120a^4b^4$$

Problem 1.2: Find the middle term in $(x + 2y)^6$.

Solution

$n = 6$ (even), middle term = $T_4$ ($r = 3$)

$$T_4 = \binom{6}{3}(x)^3(2y)^3 = 20 \times x^3 \times 8y^3 = 160x^3y^3$$

Problem 1.3: Write the general term of $\left(x - \frac{1}{x}\right)^{12}$.

Solution$$T_{r+1} = \binom{12}{r}(x)^{12-r}\left(-\frac{1}{x}\right)^r = \binom{12}{r}(-1)^r x^{12-2r}$$

Level 2: JEE Main Advanced

Problem 2.1: Find the term independent of $x$ in $\left(x^2 - \frac{3}{x}\right)^9$.

Solution

General term: $T_{r+1} = \binom{9}{r}(x^2)^{9-r}\left(-\frac{3}{x}\right)^r = \binom{9}{r}(-3)^r x^{18-3r}$

For independence: $18 - 3r = 0 \Rightarrow r = 6$

$$T_7 = \binom{9}{6}(-3)^6 = 84 \times 729 = 61236$$

Problem 2.2: If the coefficients of $(r-1)^{\text{th}}$, $r^{\text{th}}$, and $(r+1)^{\text{th}}$ terms in $(1+x)^{20}$ are in AP, find $r$.

Solution

Coefficients are: $\binom{20}{r-2}$, $\binom{20}{r-1}$, $\binom{20}{r}$

In AP: $2\binom{20}{r-1} = \binom{20}{r-2} + \binom{20}{r}$

Using $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$:

$2\binom{20}{r-1} = \binom{20}{r-2} + \binom{20}{r}$

After calculation: $r = 7$ or $r = 14$

Problem 2.3: Find the ratio of the 5th term to the 4th term in $(3x + 2y)^{10}$.

Solution$$\frac{T_5}{T_4} = \frac{\binom{10}{4}(3x)^6(2y)^4}{\binom{10}{3}(3x)^7(2y)^3}$$ $$= \frac{\binom{10}{4}}{\binom{10}{3}} \times \frac{(3x)^6}{(3x)^7} \times \frac{(2y)^4}{(2y)^3}$$ $$= \frac{10!/(4!6!)}{10!/(3!7!)} \times \frac{1}{3x} \times 2y$$ $$= \frac{7}{4} \times \frac{2y}{3x} = \frac{7y}{6x}$$

Level 3: JEE Advanced

Problem 3.1: Find the greatest term numerically in the expansion of $(3 - 5x)^{15}$ when $x = 1/5$.

Solution

For greatest term, use: $\frac{T_{r+1}}{T_r} = \frac{(n-r+1)|b|}{r|a|}$

Here $a = 3$, $b = -5x = -1$, $n = 15$

$\frac{T_{r+1}}{T_r} = \frac{(16-r) \times 1}{r \times 3} = \frac{16-r}{3r}$

For greatest term: $\frac{16-r}{3r} \geq 1$ and $\frac{16-(r+1)}{3(r+1)} < 1$

From first: $16 - r \geq 3r \Rightarrow r \leq 4$

So greatest term is $T_5$ or check ratio.

Calculate $T_4$ and $T_5$ and compare numerically.

Problem 3.2: If the 3rd term in $(x^{\alpha} + x^{-\beta})^n$ is independent of $x$, express $\alpha$ in terms of $\beta$ and $n$.

Solution

3rd term: $T_3 = T_{2+1}$, so $r = 2$

$$T_3 = \binom{n}{2}(x^{\alpha})^{n-2}(x^{-\beta})^2 = \binom{n}{2}x^{\alpha(n-2)-2\beta}$$

For independence: $\alpha(n-2) - 2\beta = 0$

$$\alpha = \frac{2\beta}{n-2}$$

Problem 3.3: Show that the middle term in $(1 + x)^{2n}$ is $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} \cdot 2^n \cdot x^n$.

Solution

Middle term when power is $2n$ (even): $T_{n+1}$ (where $r = n$)

$$T_{n+1} = \binom{2n}{n}x^n = \frac{(2n)!}{n! \cdot n!}x^n$$ $$= \frac{2n \cdot (2n-1) \cdot (2n-2) \cdots 2 \cdot 1}{n! \cdot n!}x^n$$ $$= \frac{[2 \cdot 4 \cdot 6 \cdots (2n)] \cdot [1 \cdot 3 \cdot 5 \cdots (2n-1)]}{(n!)^2}x^n$$ $$= \frac{2^n \cdot n! \cdot [1 \cdot 3 \cdot 5 \cdots (2n-1)]}{(n!)^2}x^n$$ $$= \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} \cdot 2^n \cdot x^n$$

Important Results Summary

PropertyFormula
General Term$T_{r+1} = \binom{n}{r}a^{n-r}b^r$
Middle Term ($n$ even)$T_{\frac{n}{2}+1}$
Middle Terms ($n$ odd)$T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$
Term number from end$T_r$ from end = $T_{n-r+2}$ from start
Greatest coefficient$\binom{n}{n/2}$ if $n$ even, $\binom{n}{(n-1)/2}$ or $\binom{n}{(n+1)/2}$ if $n$ odd

Cross-References

Quick Revision Checklist

  • General term formula memorized
  • Can find middle term for even and odd $n$
  • Know how to find term independent of variable
  • Understand relationship between $r$ and term number
  • Can handle fractional/negative powers
  • Practice finding specific coefficients

Last updated: October 18, 2025