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Binomial Theorem — General Term, Middle Term & Greatest Term

Master Binomial Theorem for JEE Main and Board Exams. Covers binomial expansion formula general term T(r+1)=C(n,r)a^(n-r)b^r, finding specific term given position, term independent of x by setting power to zero, middle term n even one middle term n odd two middle terms, greatest term using ratio test r≤(n+1)|x|/(1+|x|), standard results sum of coefficients, Pascal triangle, with fully verified JEE Board practice questions.

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The Binomial Theorem provides a compact formula for expanding powers of a binomial — expressions of the form (a + b)ⁿ — without actually multiplying out n times. What would take pages of algebra for (x + 2)¹⁵ becomes a single formula. The theorem is one of the oldest and most beautiful results in algebra, connecting combinatorics (binomial coefficients) to polynomial expansion. For JEE Main and Board exams, three question types dominate: finding a specific term (the General Term), identifying the Middle Term(s), finding the term independent of x, and locating the Greatest Term. Every one of these reduces to the same General Term formula — master it once and all these become mechanical.

1. Binomial Theorem — Statement

For any positive integer n and real numbers a, b:

$(a + b)^{n} = \sum_{r = 0}^{n} (\binom{n}{r}) a^{n - r} b^{r} = (\binom{n}{0}) a^{n} + (\binom{n}{1}) a^{n - 1} b + \dots + (\binom{n}{n}) b^{n}$

Where $(\binom{n}{r}) =^{n} C_{r} = \frac{n!}{r! (n - r)!}$ are the binomial coefficients.

Pascal's Triangle — First 6 Rows

n	Coefficients	Sum
0	1	1
1	1 1	2
2	1 2 1	4
3	1 3 3 1	8
4	1 4 6 4 1	16
5	1 5 10 10 5 1	32

Sum of coefficients in (a+b)ⁿ = 2ⁿ. This follows by putting a = b = 1.

2. General Term (T_{r+1})

The (r+1)th term in the expansion of (a+b)ⁿ is:

$T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r} where r = 0, 1, 2, \dots, n$

Total number of terms in the expansion = n + 1

Key pattern: In T_{r+1}, the power of a is (n−r) and the power of b is r. They always sum to n.

Finding a Specific Term — Worked Examples

Example 1: Find T₄ in the expansion of (x + 2)⁸.

T₄ = T_{3+1}: r = 3, n = 8, a = x, b = 2

T₄ = C(8,3) · x⁵ · 2³ = 56 · 8 · x⁵ = 448x⁵

Example 2: Find T₄ in the expansion of (2x − 1/x)¹⁰.

r = 3, n = 10, a = 2x, b = −1/x

T₄ = C(10,3) · (2x)⁷ · (−1/x)³ = 120 · 128 · x⁷ · (−1) · x⁻³ = −15360 · x⁴ = −15360x⁴

3. Term Independent of x (Constant Term)

To find the term not containing x: write T_{r+1} with all powers of x, set the total power of x equal to zero, solve for r, then substitute.

Step-by-Step Method

Write T_{r+1} = C(n,r) · (first part)^(n−r) · (second part)^r
Collect all x terms and write the power of x as a function of r: power(x) = f(r)
Set f(r) = 0 and solve for r (must be a non-negative integer)
Substitute r back to find the coefficient

Worked Examples

Example 1: Find the term independent of x in (x + 1/x²)⁹.

T_{r+1} = C(9,r) · x^(9−r) · (x⁻²)^r = C(9,r) · x^(9−r−2r) = C(9,r) · x^(9−3r)

Set 9 − 3r = 0 → r = 3

T₄ = C(9,3) = 84

Example 2: Find the term independent of x in (2x² − 1/x)¹².

T_{r+1} = C(12,r) · (2x²)^(12−r) · (−1/x)^r = C(12,r) · 2^(12−r) · x^(24−2r) · (−1)^r · x^(−r)

Power of x = 24 − 2r − r = 24 − 3r. Set = 0 → r = 8

T₉ = C(12,8) · 2⁴ · (−1)⁸ = 495 · 16 · 1 = 7920

4. Middle Term(s)

$n even \Rightarrow 1 middle term: T_{\frac{n}{2} + 1} n odd \Rightarrow 2 middle terms: T_{\frac{n + 1}{2}} and T_{\frac{n + 3}{2}}$

Total terms = n + 1. If n+1 is odd (n even), there is one middle term. If n+1 is even (n odd), there are two middle terms.

Examples

(a+b)¹²: n = 12 (even) → 13 terms → Middle term = T₇ (r = 6)

T₇ = C(12,6) · a⁶ · b⁶ = 924 a⁶b⁶

(a+b)⁹: n = 9 (odd) → 10 terms → Two middle terms: T₅ and T₆

5. Greatest Term in the Expansion

The numerically greatest term in (1+x)ⁿ is found using the ratio test:

$\frac{| T_{r + 1} |}{| T_{r} |} = \frac{(n - r + 1) | x |}{r}$

T_{r+1} ≥ T_r when this ratio ≥ 1, i.e., when:

$r \leq \frac{(n + 1) | x |}{1 + | x |}$

Let m = (n+1)|x|/(1+|x|). The greatest term is T_{⌊m⌋+1} if m is not an integer; if m is an integer, T_m and T_{m+1} are equally greatest.

Worked Example

Find the greatest term in (1+2)⁸ (i.e., x = 2, n = 8):

m = (8+1)×2/(1+2) = 18/3 = 6 (integer) → T₆ and T₇ are equally greatest

T₇ = C(8,6) · 2⁶ = 28 · 64 = 1792 (both T₆ and T₇ equal 1792)

6. Key Standard Results

Result	Formula	Obtained by
Sum of all coefficients	2ⁿ	Put a = b = 1 in (a+b)ⁿ
Sum of even-position coefficients	2^(n-1)	Average of (1+1)ⁿ and (1−1)ⁿ
Sum of odd-position coefficients	2^(n-1)	Same as above
C₀ − C₁ + C₂ − ··· = 0	0	Put x = −1 in (1+x)ⁿ
Expansion of (1+x)ⁿ at x=1	C₀+C₁+C₂+···+Cₙ = 2ⁿ	Put x = 1

4-Step Universal Method for Any Binomial Term Problem

Step	Action
1. Identify	n, a (first term), b (second term including sign)
2. Write T_{r+1}	T_{r+1} = C(n,r) · a^(n−r) · b^r
3. Set condition	For specific term: set r = given value. For x-independent: set power of x = 0. For middle: r = n/2 or (n±1)/2
4. Compute	Substitute r, compute C(n,r) and remaining powers; include sign from (−b)^r

JEE/Board Power Tips — Avoid These Common Mistakes

T_{r+1} starts at r = 0, NOT r = 1. T₁ corresponds to r = 0, T₂ to r = 1, and so on. Always write T_{r+1} = C(n,r)·aⁿ⁻ʳ·bʳ and substitute r = (required term number − 1).
The sign of the second term goes into b. For (2x − 1/x)¹⁰, treat b = −1/x (with the negative sign). Then (−1/x)^r automatically gives the correct sign for each term.
Middle term: n even → 1 middle term (T_{n/2+1}); n odd → 2 middle terms (T_{(n+1)/2} and T_{(n+3)/2}). The easy check: n + 1 total terms; if odd count, one middle; if even count, two middles.
For term independent of x: set power of x = 0 and check r is a non-negative integer. If r is not an integer, there is no term independent of x in that expansion. Always verify.
Sum of coefficients ≠ sum of binomial coefficients. Sum of coefficients of (2x + 3)ⁿ is obtained by putting x = 1 → (2+3)ⁿ = 5ⁿ, not 2ⁿ. Sum of binomial coefficients only = 2ⁿ when the expansion is (1+x)ⁿ with coefficient 1.

Practice Questions

Q1 (Board / JEE Main): Find the $5^{th}$ term in the binomial expansion of $(3 x - 2 y)^{8}$ .

Explanation:

The general term in the expansion of $(a + b)^{n}$ is given by $T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r}$ .

To find the $5^{th}$ term ( $T_{5}$ ), we set $r = 4$ .
Here, $n = 8$ , $a = 3 x$ , and $b = - 2 y$ .

Substitute the values into the general formula:
$T_{5} = (\binom{8}{4}) \cdot (3 x)^{8 - 4} \cdot (- 2 y)^{4}$
$T_{5} = (\binom{8}{4}) \cdot (3 x)^{4} \cdot (- 2 y)^{4}$

Calculate the components:
$(\binom{8}{4}) = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$
$(3 x)^{4} = 81 x^{4}$
$(- 2 y)^{4} = 16 y^{4}$

Multiply them together:
$T_{5} = 70 \cdot 81 x^{4} \cdot 16 y^{4}$
$T_{5} = 90720 x^{4} y^{4}$

Q2 (JEE Main): Find the term strictly independent of $x$ (the constant term) in the expansion of ${(\frac{x}{3} - \frac{3}{x^{2}})}^{9}$ .

Explanation:

Write the general term $T_{r + 1}$ for the expansion:
$T_{r + 1} = (\binom{9}{r}) \cdot {(\frac{x}{3})}^{9 - r} \cdot {(- \frac{3}{x^{2}})}^{r}$

Separate the constants from the variable $x$ :
$T_{r + 1} = (\binom{9}{r}) \cdot \frac{x^{9 - r}}{3^{9 - r}} \cdot (- 3)^{r} \cdot x^{- 2 r}$
$T_{r + 1} = (\binom{9}{r}) \cdot \frac{(- 3)^{r}}{3^{9 - r}} \cdot x^{(9 - r) - 2 r}$
$T_{r + 1} = (\binom{9}{r}) \cdot \frac{(- 3)^{r}}{3^{9 - r}} \cdot x^{9 - 3 r}$

For the term to be strictly independent of $x$ , the overall power of $x$ must be zero:
$9 - 3 r = 0 ⟹ 3 r = 9 ⟹ r = 3$

Now, substitute $r = 3$ back into the separated general term to find its value ( $T_{4}$ ):
$T_{4} = (\binom{9}{3}) \cdot \frac{(- 3)^{3}}{3^{9 - 3}} \cdot x^{0}$
$T_{4} = 84 \cdot \frac{- 27}{3^{6}} = 84 \cdot \frac{- 27}{729}$

Simplify the fraction:
$T_{4} = 84 \cdot (- \frac{1}{27}) = - \frac{84}{27} = - \frac{28}{9}$

Q3 (Board): Find the middle term(s) in the expansion of ${(\frac{x}{2} + \frac{2}{x})}^{10}$ .

Explanation:

Since the index $n = 10$ is an even number, the total number of terms in the expansion is $n + 1 = 11$ (an odd number). Therefore, there is only exactly one middle term.

Position of the middle term = ${(\frac{n}{2} + 1)}^{th}$ term = $(\frac{10}{2} + 1) = 6^{th}$ term.
So, we need to calculate $T_{6}$ .

Using the general term formula $T_{r + 1}$ with $r = 5$ :
$T_{6} = (\binom{10}{5}) \cdot {(\frac{x}{2})}^{5} \cdot {(\frac{2}{x})}^{5}$

Notice that the variable $x$ and the constant $2$ cancel out completely:
$T_{6} = 252 \cdot (\frac{x^{5}}{32}) \cdot (\frac{32}{x^{5}})$
$T_{6} = 252 \cdot 1 = 252$

Q4 (MCQ — 1 mark): The coefficient of $x^{5}$ in the expansion of $(1 + x)^{10}$ is:

A) 210
B) 252
C) 120
D) 45

Answer: B) 252.

Explanation: For the standard expansion of $(1 + x)^{n}$ , the coefficient of $x^{r}$ is simply given by $(\binom{n}{r})$ .
Here, $n = 10$ and $r = 5$ .
Coefficient = $(\binom{10}{5}) = \frac{10!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = \frac{30240}{120} = 252$ .

Q5 (JEE Main): If the respective coefficients of $x^{7}$ and $x^{8}$ in the expansion of ${(2 + \frac{x}{3})}^{n}$ are equal, mathematically determine the value of $n$ .

Explanation:

First, write the general term $T_{r + 1}$ for the expansion:
$T_{r + 1} = (\binom{n}{r}) \cdot 2^{n - r} \cdot {(\frac{x}{3})}^{r} = (\binom{n}{r}) \cdot \frac{2^{n - r}}{3^{r}} \cdot x^{r}$

From this, we extract the coefficient for any power $r$ : $(\binom{n}{r}) \cdot \frac{2^{n - r}}{3^{r}}$

Coefficient of $x^{7}$ (where $r = 7$ ): $(\binom{n}{7}) \cdot \frac{2^{n - 7}}{3^{7}}$
Coefficient of $x^{8}$ (where $r = 8$ ): $(\binom{n}{8}) \cdot \frac{2^{n - 8}}{3^{8}}$

We are given that these two coefficients are exactly equal:
$(\binom{n}{7}) \cdot \frac{2^{n - 7}}{3^{7}} = (\binom{n}{8}) \cdot \frac{2^{n - 8}}{3^{8}}$

Rearrange the terms to group combinations and exponents:
$\frac{(\binom{n}{8})}{(\binom{n}{7})} = (\frac{2^{n - 7}}{2^{n - 8}}) \cdot (\frac{3^{8}}{3^{7}})$
$\frac{(\binom{n}{8})}{(\binom{n}{7})} = (2^{1}) \cdot (3^{1}) = 6$

Now, expand the ratio of the binomial coefficients using the property $\frac{(\binom{n}{r})}{(\binom{n}{r - 1})} = \frac{n - r + 1}{r}$ :
$\frac{(\binom{n}{8})}{(\binom{n}{7})} = \frac{n - 8 + 1}{8} = \frac{n - 7}{8}$

Set this equal to 6 and solve for $n$ :
$\frac{n - 7}{8} = 6$
$n - 7 = 48 ⟹ n = 55$

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Binomial Theorem — General Term, Middle Term & Greatest Term

1. Binomial Theorem — Statement

Pascal's Triangle — First 6 Rows

2. General Term (T_{r+1})

Finding a Specific Term — Worked Examples

3. Term Independent of x (Constant Term)

Step-by-Step Method

Worked Examples

4. Middle Term(s)

Examples

5. Greatest Term in the Expansion

Worked Example

6. Key Standard Results

4-Step Universal Method for Any Binomial Term Problem

Practice Questions

Related Study Material

Practice Questions

Topic Information

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Track Your Learning

Study Tips

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