Mathematics · Ch 6 — Binomial Theorem

Some Special Cases

6.2.2

Some Special Cases

Concept First: Why Special Cases Matter

The binomial theorem gives us a general expansion for $(a+b)^n$ , but in practice we rarely work with generic $a$ and $b$ . Most problems involve specific patterns — a difference like $(x-y)^n$ , a sum like $(1+x)^n$ , or a difference like $(1-x)^n$ . Each of these patterns simplifies the general expansion into a form with alternating signs or just positive terms, and each leads to important identities that appear repeatedly in competitive exams.

The key insight is simple: substitute particular values for $a$ and $b$ into the general formula, then simplify. The three cases below are the ones the NCERT textbook treats explicitly, and they form the backbone of almost every binomial theorem problem you will face.

Case (i): $(x - y)^n$ — The Alternating-Sign Expansion

Take $a = x$ and $b = -y$ in the general expansion:

(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n

Substituting gives:

(x - y)^n = [x + (-y)]^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}(-y) + \binom{n}{2}x^{n-2}(-y)^2 + \binom{n}{3}x^{n-3}(-y)^3 + \cdots + \binom{n}{n}(-y)^n

Now simplify each term. Since $(-y)^k = (-1)^k y^k$ , we get:

(x - y)^n = \binom{n}{0}x^n - \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 - \binom{n}{3}x^{n-3}y^3 + \cdots + (-1)^n\binom{n}{n}y^n

(x - y)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{n-k} y^k

The signs alternate: $+$ , $-$ , $+$ , $-$ , $\dots$ , ending with $(-1)^n$ .

Watch out

The last term is $(-1)^n \binom{n}{n} y^n$ , not $(-1)^n \binom{n}{n} x^0 y^n$ — the $x^0$ factor is 1 and is usually omitted. Students often forget the sign on the last term.

Worked Example: $(x - 2y)^5$

Using the formula above with $n=5$ , $x$ as $x$ , and $y$ replaced by $2y$ :

(x - 2y)^5 = \binom{5}{0}x^5 - \binom{5}{1}x^4(2y) + \binom{5}{2}x^3(2y)^2 - \binom{5}{3}x^2(2y)^3 + \binom{5}{4}x(2y)^4 - \binom{5}{5}(2y)^5

Compute each coefficient:

$\binom{5}{0}=1$ , term: $x^5$
$\binom{5}{1}=5$ , term: $-5 \cdot x^4 \cdot 2y = -10x^4y$
$\binom{5}{2}=10$ , term: $10 \cdot x^3 \cdot 4y^2 = 40x^3y^2$
$\binom{5}{3}=10$ , term: $-10 \cdot x^2 \cdot 8y^3 = -80x^2y^3$
$\binom{5}{4}=5$ , term: $5 \cdot x \cdot 16y^4 = 80xy^4$
$\binom{5}{5}=1$ , term: $-1 \cdot 32y^5 = -32y^5$

So:

(x - 2y)^5 = x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5

Case (ii): $(1 + x)^n$ — The All-Positive Expansion

Take $a = 1$ , $b = x$ in the general formula:

(1 + x)^n = \binom{n}{0}(1)^n + \binom{n}{1}(1)^{n-1}x + \binom{n}{2}(1)^{n-2}x^2 + \cdots + \binom{n}{n}x^n

Since $(1)^k = 1$ for any $k$ , this simplifies to:

(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \cdots + \binom{n}{n}x^n

(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k

All terms are positive. This is the most frequently used form of the binomial theorem.

A Crucial Special Case: $x = 1$

Substitute $x = 1$ into the expansion:

(1 + 1)^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n}

The left side is $2^n$ . Therefore:

Important

$2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n}$

This identity tells us that the sum of all binomial coefficients for a given $n$ equals $2^n$ . It is one of the most frequently tested results in competitive exams.

Case (iii): $(1 - x)^n$ — The Alternating-Sign Expansion (Again)

Take $a = 1$ , $b = -x$ in the general formula:

(1 - x)^n = \binom{n}{0}(1)^n + \binom{n}{1}(1)^{n-1}(-x) + \binom{n}{2}(1)^{n-2}(-x)^2 + \cdots + \binom{n}{n}(-x)^n

Since $(-x)^k = (-1)^k x^k$ , we get:

$(1 - x)^n = \binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^2 - \binom{n}{3}x^3 + \cdots + (-1)^n\binom{n}{n}x^n$ …

Start your 14-day free trial to unlock the full solution →

← Previous · 6.2.1Binomial Theorem for Any Positive Integer n Next · 6.2.2 →Example 1 — Expand (x² + (3)/(x))⁴, x ≠ 0.

Some Special Cases