Binomial Theorem

The Binomial Theorem provides a formula for expanding expressions of the form `(x + a)^n` where `n` is a positive integer. It is a generalization of familiar algebraic identities like `(a+b)^2`. The terms in the expansion are built using binomial coefficients, which are identical to the combinations `nCr` from the chapter on Permutations and Combinations. A solid understanding of combinations is therefore essential before starting this topic.

Students are expected to learn how to write the complete expansion, find a specific term (the general term), locate the middle term(s), and find terms with specific properties, like the term independent of a variable. The chapter also explores the many properties of binomial coefficients, which are used to solve problems involving sums of series. The difficulty usually rises with problems that ask for the numerically greatest term in an expansion or involve complex summations. For JEE Advanced, the theorem is extended to handle any rational exponent `n`, which results in an infinite series.

Formulas for Binomial Theorem

Binomial expansion for a positive integer index: $(x+a)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} a^r$ \binom{n}{r} is the binomial coefficient, equal to \frac{n!}{r!(n-r)!}.
General term in the expansion of (x+a)^n: $T_{r+1} = \binom{n}{r} x^{n-r} a^r$ This is the (r+1)-th term, as the index r starts from 0.
Middle term(s) in the expansion of (x+a)^n: $\text{The } (\frac{n}{2}+1)\text{-th term, if n is even.}$ If n is odd, there are two middle terms: the (\frac{n+1}{2})-th and (\frac{n+3}{2})-th terms.
Sum of all binomial coefficients: $\sum_{r=0}^{n} \binom{n}{r} = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$ This is the sum of coefficients in the expansion of (1+x)^n.
Alternating sum of binomial coefficients: $\sum_{r=0}^{n} (-1)^r \binom{n}{r} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots = 0$ This holds for any integer n > 0.
Sum of even and odd indexed coefficients: $\binom{n}{0} + \binom{n}{2} + \dots = \binom{n}{1} + \binom{n}{3} + \dots = 2^{n-1}$ Each sum is equal to half the total sum of coefficients.
Ratio of consecutive binomial coefficients: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$ Useful for comparing terms and finding the greatest coefficient.
Binomial theorem for any rational index: $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$ This infinite series expansion is valid only when |x| < 1.

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JEE Advanced 2023

Let a and b be two non-zero real numbers. If the coefficient of $x^{5}$ in the expansion of (ax $^{2} + 70/(27$ bx $))^{4}$ is equal to the coefficient of $x^{-5}$ in the expansion of (ax $- 1/($ bx $^{2}))^{7},$ then the value of 2b is

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$a, b \neq 0, \quad a, b \in \mathbb{R}$ $\left(ax^{2} + \dfrac{70}{27bx}\right)^{4}$ $\left(ax - \dfrac{1}{bx^{2}}\right)^{7}$ $\text{Coeff of } x^{5} \text{ in } (1) = \text{Coeff of } x^{-5} \text{ in } (2)$ $\text{Find } 2b$ $\text{Plan: write } T_{r+1} \text{ for each, match the power of } x$ $\text{then equate the two coefficients}$ $(X+Y)^{n} = \sum_{r=0}^{n} \binom{n}{r} X^{n-r} Y^{r}$ $T_{r+1} = \binom{n}{r} X^{n-r} Y^{r}$ $T_{k+1} = \binom{4}{k} (ax^{2})^{4-k} \left(\dfrac{70}{27bx}\right)^{k}$ $= \binom{4}{k} a^{4-k}\!\left(\dfrac{70}{27b}\right)^{k} x^{\,8-3k}$ $8-3k = 5 \ \Rightarrow\ k = 1$ $\text{Coeff of } x^{5} = \binom{4}{1} a^{3}\,\dfrac{70}{27b} = \dfrac{280\,a^{3}}{27b}$ $T_{j+1} = \binom{7}{j} (ax)^{7-j} \left(-\dfrac{1}{bx^{2}}\right)^{j}$ $= \binom{7}{j} (-1)^{j} a^{7-j} b^{-j}\, x^{\,7-3j}$ $7-3j = -5 \ \Rightarrow\ j = 4$ $\text{Coeff of } x^{-5} = \binom{7}{4} a^{3} b^{-4} = \dfrac{35\,a^{3}}{b^{4}}$ $\dfrac{280\,a^{3}}{27b} = \dfrac{35\,a^{3}}{b^{4}}$ $\Rightarrow \dfrac{280}{27b} = \dfrac{35}{b^{4}} \quad (a \neq 0)$ $280\,b^{4} = 35 \cdot 27\,b$ $\Rightarrow 280\,b^{3} = 945 \ \Rightarrow\ b^{3} = \dfrac{945}{280} = \dfrac{27}{8}$ $b = \sqrt[3]{\dfrac{27}{8}} = \dfrac{3}{2}$ $2b = 3$ $\text{Coeff of } x^{5} = \dfrac{280\,a^{3}}{27\cdot\frac{3}{2}} = \dfrac{560\,a^{3}}{81}$ $\text{Coeff of } x^{-5} = \dfrac{35\,a^{3}}{\left(\frac{3}{2}\right)^{4}} = \dfrac{560\,a^{3}}{81}$

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Past-paper questions on Binomial Theorem

JEE Advanced 8 questions