# Binomial Theorem for Negative/ Rational Index

## Binomial Theorem for Negative/ Rational Index

Let n is a rational number and x is a real number such that |x| < 1, then $${{\left( 1+x \right)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+\frac{n(n-1)(n-2)}{3!}{{x}^{3}}+……$$.

(i) The expansion of (x + a)n

$$={{a}^{n}}{{\left( 1+\frac{x}{a} \right)}^{n}}$$.

$$={{a}^{n}}\left( 1+n\times \left( \frac{x}{a} \right)+\frac{n(n-1)}{2!}{{\left( \frac{x}{a} \right)}^{2}}+\frac{n(n-1)(n-2)}{3!}{{\left( \frac{x}{a} \right)}^{3}}+….. \right)$$.

The above expansion is valid only, when |x/a|< a.

(ii) The expansion (2 + 3x)⁻⁵ up to four terms in decreasing power of x is as follows.

$${{(2+3x)}^{-5}}={{\left[ 3x\left[ 1+\frac{2}{3x} \right] \right]}^{-5}}$$.

$$=\frac{1}{243{{x}^{5}}}\left[ 1+(-5)\left( \frac{2}{3x} \right)+\frac{(-5)(-5-1)}{2!}{{\left( \frac{2}{3x} \right)}^{2}}+\frac{(-5)(-5-1)(-5-2)}{3!}{{\left( \frac{2}{3x} \right)}^{3}}+…. \right]$$.

$$=\frac{1}{243{{x}^{5}}}\left[ 1+(-5)\left( \frac{2}{3x} \right)+\frac{(-5)(-6)}{2!}{{\left( \frac{2}{3x} \right)}^{2}}+\frac{(-5)(-6)(-7)}{3!}{{\left( \frac{2}{3x} \right)}^{3}}+…. \right]$$.

$$=\frac{1}{243{{x}^{5}}}\left[ 1-\frac{10}{3x}+\frac{20}{3{{x}^{2}}}-\frac{280}{27{{x}^{3}}}+…. \right]$$.

$$=\frac{1}{243}\left[ \frac{1}{{{x}^{5}}}-\frac{10}{3x}.\frac{1}{{{x}^{5}}}+\frac{20}{3{{x}^{2}}}.\frac{1}{{{x}^{5}}}-\frac{280}{27{{x}^{3}}}.\frac{1}{{{x}^{5}}}+…. \right]$$.

Note:

• If n is a whole number, then there is no need of condition |x|<1
• In the expansion (1 + x)n

(i) If n is natural number, then there are finite number terms exist.

(ii) If n is negative/ rational number, then there are infinite number of terms exist.