Binomial Theorem for any Index - Practice Questions & MCQ

Binomial Theorem for any  index

$(1+x)^n$ for negative or fractional Index

$
(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots+\frac{n(n-1)(n-2) \ldots . .(n-r+1)}{r!} x^r \ldots \infty
$
Note:
1. If n is a negative or fractional index then the condition $|\mathrm{x}|<1$ is essential.
2. There is an infinite number of terms in the expansion of $(1+x)^n$ when $n$ is a negative or fractional index.|

If the first term is not unity and the index of the binomial is either a negative integer or a fraction, then we expand as follows:

$
\begin{aligned}
& \text { If }|x|<|a| \\
& \begin{aligned}
(x+a)^n & =\left\{a\left(1+\frac{x}{a}\right)\right\}^n=a^n\left(1+\frac{x}{a}\right)^n \\
& =a^n\left\{1+n \frac{x}{a}+\frac{n(n-1)}{2!}\left(\frac{x}{a}\right)^2+\cdots\right\} \\
& =a^n+n a^{n-1} x+\frac{n(n-1)}{2!} a^{n-2} x^2+\cdots
\end{aligned}
\end{aligned}
$
And when $|x|>|a|$

$
\begin{aligned}
(x+a)^n & =\left\{x\left(1+\frac{a}{x}\right)\right\}^n \\
& =x^n\left\{1+n \frac{a}{x}+\frac{n(n-1)}{2!}\left(\frac{a}{x}\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac{a}{x}\right)^3+\cdots\right\}
\end{aligned}
$