The Binomial Theorem and Its Special Cases

The Theorem

Key Point (Binomial Theorem): For any positive integer $n$,

$$(a+b)^n = {^nC_0}a^n + {^nC_1}a^{n-1}b + {^nC_2}a^{n-2}b^2 + \cdots + {^nC_{n-1}}a\,b^{n-1} + {^nC_n}b^n = \sum_{k=0}^{n} {^nC_k}\,a^{n-k}b^k$$

The $^nC_r$ are the binomial coefficients. The proof is by mathematical induction: $P(1)$ is immediate, and multiplying the assumed expansion of $(a+b)^k$ by one more $(a+b)$, then grouping like terms, produces exactly row $k+1$ — because the coefficients combine by the Pascal identity $^kC_r + {^kC_{r-1}} = {^{k+1}C_r}$.

A worked illustration:

$$(x+2)^6 = x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64$$

where each coefficient is $^6C_r \cdot 2^r$ — e.g. the $x^3$ term is $^6C_3\,x^3\,2^3 = 20 \cdot 8\,x^3 = 160x^3$.

[Board Tip] Bookkeeping wins marks: for index $n$ expect exactly $n+1$ terms, powers of the first quantity falling $n \to 0$, powers of the second rising $0 \to n$, and each term's indices adding to $n$. Count your terms before moving on.

Special Cases That Do the Real Work

Replace $b$ by $-y$ — signs alternate, starting positive:

$$(x-y)^n = {^nC_0}x^n - {^nC_1}x^{n-1}y + {^nC_2}x^{n-2}y^2 - \cdots + (-1)^n\,{^nC_n}y^n$$

For example: $(x - 2y)^5 = x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5$.

Set $a = 1$ — the workhorse forms:

$$(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + \cdots + {^nC_n}x^n$$

$$(1-x)^n = {^nC_0} - {^nC_1}x + {^nC_2}x^2 - \cdots + (-1)^n\,{^nC_n}x^n$$

Key Point (put $x = 1$): $\;{^nC_0} + {^nC_1} + \cdots + {^nC_n} = 2^n$ and $\;{^nC_0} - {^nC_1} + {^nC_2} - \cdots + (-1)^n\,{^nC_n} = 0$ — the two coefficient identities every exam quotes.

[JEE Tip] Adding and subtracting the two identities splits the coefficients: the even-position ones ($^nC_0 + {^nC_2} + \cdots$) and the odd-position ones ($^nC_1 + {^nC_3} + \cdots$) each sum to $2^{n-1}$.

Solved Examples

Example 1: Fractional second term

Expand $\left(x^2 + \frac{3}{x}\right)^4$, $x \ne 0$.

Solution:

Step 1 — Set up the theorem. With $a = x^2$, $b = \frac{3}{x}$, $n = 4$:

$$^4C_0(x^2)^4 + {^4C_1}(x^2)^3\frac{3}{x} + {^4C_2}(x^2)^2\frac{9}{x^2} + {^4C_3}(x^2)\frac{27}{x^3} + {^4C_4}\frac{81}{x^4}$$

Step 2 — Evaluate the coefficients. $1, 4, 6, 4, 1$ times $3^0, 3^1, 3^2, 3^3, 3^4$: $1, 12, 54, 108, 81$.

Step 3 — Combine the powers of $x$. $x^8, x^6 \cdot x^{-1} = x^5, x^4 \cdot x^{-2} = x^2, x^2 \cdot x^{-3} = x^{-1}, x^{-4}$:

$$= x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$$

Takeaway: Powers of $x$ drop by 3 each term here — the exponent walk $8, 5, 2, -1, -4$ is a quick self-check.

Example 2: Alternating signs

Expand $(1 - 2x)^5$.

Solution:

Step 1 — Use the $(1-x)$ pattern. Signs alternate starting positive:

$$(1-2x)^5 = {^5C_0} - {^5C_1}(2x) + {^5C_2}(2x)^2 - {^5C_3}(2x)^3 + {^5C_4}(2x)^4 - {^5C_5}(2x)^5$$

Step 2 — Evaluate. Coefficients $^5C_r 2^r$: $1, 10, 40, 80, 80, 32$ with alternating signs:

$$= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$$

Step 3 — Check. At $x = 1$: $1 - 10 + 40 - 80 + 80 - 32 = -1 = (1-2)^5$ ✓.

Takeaway: $(2x)^r$ contributes $2^r$ — and the $x = 1$ substitution verifies the whole line at once.

Example 3: Both terms fractional

Expand $\left(\frac{2}{x} - \frac{x}{2}\right)^5$, $x \ne 0$.

Solution:

Step 1 — Identify the parts. $a = \frac{2}{x}$, $b = -\frac{x}{2}$; odd powers of $b$ carry minus signs.

Step 2 — Write the five-row expansion.

$$\frac{32}{x^5} - 5 \cdot \frac{16}{x^4} \cdot \frac{x}{2} + 10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4} - 10 \cdot \frac{4}{x^2} \cdot \frac{x^3}{8} + 5 \cdot \frac{2}{x} \cdot \frac{x^4}{16} - \frac{x^5}{32}$$

Step 3 — Simplify term by term. $\frac{32}{x^5}, \ -\frac{40}{x^3}, \ +\frac{20}{x}, \ -5x, \ +\frac{5x^3}{8}, \ -\frac{x^5}{32}$:

$$= \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}$$

Takeaway: Powers of $x$ climb by 2 each term ($-5$ to $+5$) — another exponent walk to check against.

Example 4: A sixth-power expansion

Expand $(2x - 3)^6$.

Solution:

Step 1 — Set up with alternating signs.

$$^6C_0(2x)^6 - {^6C_1}(2x)^5(3) + {^6C_2}(2x)^4(9) - {^6C_3}(2x)^3(27) + {^6C_4}(2x)^2(81) - {^6C_5}(2x)(243) + {^6C_6}(729)$$

Step 2 — Evaluate each coefficient. $64, \ 6 \cdot 32 \cdot 3 = 576, \ 15 \cdot 16 \cdot 9 = 2160, \ 20 \cdot 8 \cdot 27 = 4320, \ 15 \cdot 4 \cdot 81 = 4860, \ 6 \cdot 2 \cdot 243 = 2916, \ 729$.

Step 3 — Assemble.

$$= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$$

Step 4 — Check. At $x = 1$: $64 - 576 + 2160 - 4320 + 4860 - 2916 + 729 = 1 = (2-3)^6$ ✓.

Takeaway: Seven terms for index 6, signs strictly alternating — count both before simplifying.

Example 5: Fractional first term

Expand $\left(\frac{x}{3} + \frac{1}{x}\right)^5$, $x \ne 0$.

Solution:

Step 1 — Row 5 with $a = \frac{x}{3}$, $b = \frac{1}{x}$.

$$\left(\frac{x}{3}\right)^5 + 5\left(\frac{x}{3}\right)^4\frac{1}{x} + 10\left(\frac{x}{3}\right)^3\frac{1}{x^2} + 10\left(\frac{x}{3}\right)^2\frac{1}{x^3} + 5\left(\frac{x}{3}\right)\frac{1}{x^4} + \frac{1}{x^5}$$

Step 2 — Simplify the powers of 3 and $x$.

$$= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$$

Takeaway: Denominators run $3^5, 3^4, \ldots$ against rising powers of $\frac{1}{x}$ — all signs positive since both terms are positive.

Example 6: A symmetric expansion

Expand $\left(x + \frac{1}{x}\right)^6$, $x \ne 0$.

Solution:

Step 1 — Row 6 and the exponent walk. Coefficients $1, 6, 15, 20, 15, 6, 1$; powers of $x$ run $6, 4, 2, 0, -2, -4, -6$ (each step trades an $x$ for an $x^{-1}$).

Step 2 — Write it out.

$$x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$$

Step 3 — Note the constant. The middle term $^6C_3\,x^3 x^{-3} = 20$ is where the powers cancel exactly.

Takeaway: In $\left(x + \frac{1}{x}\right)^n$ the exponents fall by 2 per term — the constant term exists only when $n$ is even.

Example 7: $(x - 2y)^5$ line by line

Expand $(x - 2y)^5$.

Solution:

Step 1 — Row 5 with alternating signs.

$$x^5 - 5x^4(2y) + 10x^3(4y^2) - 10x^2(8y^3) + 5x(16y^4) - 32y^5$$

Step 2 — Simplify.

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

Step 3 — Check. At $x = 2, y = 1$: $32 - 160 + 320 - 320 + 160 - 32 = 0 = (2-2)^5$ ✓.

Takeaway: $(2y)^r$ compounds to $2^r y^r$; the signs alternate because the second quantity is negative.

Example 8: The sigma form, read off

Write $\sum_{k=0}^{4} {^4C_k}\,3^{4-k}x^k$ as a power of a binomial.

Solution:

Step 1 — Pattern-match. The theorem reads $\sum {^nC_k}a^{n-k}b^k = (a+b)^n$.

Step 2 — Identify the parts. Here $n = 4$, $a = 3$ (falling powers), $b = x$ (rising powers).

Step 3 — Conclude. The sum is $(3 + x)^4$.

Takeaway: Falling-power base is $a$, rising-power base is $b$ — read sigma forms straight back into $(a+b)^n$.

Example 9: Coefficient identities in action

Evaluate (i) $^{10}C_0 + {^{10}C_1} + \cdots + {^{10}C_{10}}$ (ii) $^{10}C_0 - {^{10}C_1} + \cdots + {^{10}C_{10}}$ (iii) $^{10}C_1 + {^{10}C_3} + \cdots + {^{10}C_9}$.

Solution:

Step 1 — (i) Full sum. Put $x = 1$ in $(1+x)^{10}$: $2^{10} = 1024$.

Step 2 — (ii) Alternating sum. Put $x = 1$ in $(1-x)^{10}$: $(1-1)^{10} = 0$.

Step 3 — (iii) Odd positions. Subtracting (ii) from (i) doubles the odd-position sum: $\frac{1024 - 0}{2} = 2^9 = 512$.

Takeaway: Full sum $2^n$, alternating sum $0$ — adding/subtracting splits even and odd positions into $2^{n-1}$ each.

Example 10: A disguised expansion

Prove that $\sum_{r=0}^{n} 3^r\,{^nC_r} = 4^n$.

Solution:

Step 1 — Start from the workhorse form. $(1 + x)^n = \sum_{r} {^nC_r}x^r$ for every real $x$.

Step 2 — Choose $x = 3$. The right side becomes $\sum {^nC_r}3^r$ — exactly the given sum.

Step 3 — Evaluate the left side. $(1+3)^n = 4^n$. ∎

Takeaway: Sums of the shape $\sum {^nC_r}\,t^r$ are $(1+t)^n$ in disguise — substitute, don't compute.