Binomial Theorem
2.0 Binomial Theorem (for positive integral index)
2.0 Binomial Theorem (for positive integral index)
A formulae by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.
If $n$ is a positive integer and $x,y \in C$ then $$\begin{equation} \begin{aligned} {(x + y)^n} = {}^n{C_0}{x^{n - 0}}{y^0} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + ... + {}^n{C_r}{x^{n - r}}{y^r} + ... + {}^n{C_{n - 1}}x{y^{n - 1}} + {}^n{C_n}{x^0}{y^n} \\ \Rightarrow {(x + y)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^{n - r}}{y^r}} \\\end{aligned} \end{equation} $$
where $^n{C_0}{,^n}{C_1}{,^n}{C_2}{...^n}{C_n}$ are binomial coefficients.
The above equation can also be expressed by replacing $y$ with $a$ as $$\begin{equation} \begin{aligned} {(a + x)^n}{ = ^n}{C_0}{a^{n - 0}}{x^0}{ + ^n}{C_1}{a^{n - 1}}{x^1}{ + ^n}{C_2}{a^{n - 2}}{x^2} + ...{ + ^n}{C_r}{a^{n - r}}{x^r} + ...{ + ^n}{C_{n - 1}}a{x^{n - 1}}{ + ^n}{C_n}{a^0}{x^n}\quad ...(1) \\ \Rightarrow {(a + x)^n} = \sum\limits_{r = 0}^n {^n{C_r}{a^{n - r}}{x^r}} \\\end{aligned} \end{equation} $$
If we replace $x$ by $-x$ in equation $(1)$, we get $$\begin{equation} \begin{aligned} {(a - x)^n}{ = ^n}{C_0}{a^{n - 0}}{( - x)^0}{ - ^n}{C_1}{a^{n - 1}}{x^1}{ + ^n}{C_2}{a^{n - 2}}{x^2} - ...{ + ^n}{C_r}{a^{n - r}}{( - x)^r} + ...{ + ^n}{C_{n - 1}}a{( - x)^{n - 1}}{ + ^n}{C_n}{a^0}{( - x)^n}\quad ...(2) \\ \Rightarrow {(a - x)^n} = \sum\limits_{r = 0}^n {^n{C_r}{a^{n - r}}{{( - x)}^r}} \\\end{aligned} \end{equation} $$
Note: Total number of terms in the expansion is $=n+1$.
If there are more than two terms in an algebraic expression, then consider the constant term as first term and remaining other terms as second term and follow the same procedure explained above i.e., $${(x + y + z)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}}(y + z) + {}^n{C_2}{x^{n - 2}}{(y + z)^2} + ... + {}^n{C_n}{(y + z)^n}$$
Note: Total number of terms in the expansion is $\frac{{(n + 1)(n + 2)}}{2}$ provided $x \ne y \ne z$.
Question 1. Expand the following:
(i) ${(2 + 3x)^5}$
(ii) ${(3 - 5x)^4}$
Solution: (i) $${(2 + 3x)^5} = {}^5{C_0}{.2^5} + {}^5{C_1}{.2^4}.(3x) + {}^5{C_2}{.2^3}.{(3x)^2} + {}^5{C_3}{.2^2}.{(3x)^3} + {}^5{C_4}.2.{(3x)^4} + {}^5{C_5}{.2^0}.{(3x)^5}$$
(ii) $${(3 - 5x)^4} = {}^4{C_0}{.3^4} - {}^4{C_1}{.3^3}.(5x) + {}^4{C_2}{.3^2}.{(5x)^2} - {}^4{C_3}.3.{(5x)^3} + {}^4{C_4}.{(5x)^4}$$