JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Binomial Theorem for any Index

Statement :

\[{{(1+x)}^{n}}=1+nx+\frac{n(n-1){{x}^{2}}}{2!}+\frac{n(n-1)\,(n-2)}{3!}{{x}^{3}}+....\]\[+\frac{n(n-1)\,......(n-r+1)}{r!}{{x}^{r}}+...\text{terms up to }\infty \] 

when \[n\] is a negative integer or a fraction, where \[-1<x<1\], otherwise expansion will not be possible.

If first term is not 1, then make first term unity in the following way, \[{{(x+y)}^{n}}={{x}^{n}}{{\left[ 1+\frac{y}{x} \right]}^{n}}\],  if \[\left| \,\frac{y}{x}\, \right|<1\].

General term : \[{{T}_{r+1}}=\frac{n(n-1)(n-2)......(n-r+1)}{r!}{{x}^{r}}\]

Some important expansions

(i) \[{{(1+x)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+\] \[.......+\frac{n(n-1)\,(n-2)......(n-r+1)}{r!}{{x}^{r}}+......\]

(ii) \[{{(1-x)}^{n}}=1-nx+\frac{n(n-1)}{2!}{{x}^{2}}-.......\] \[+\frac{n(n-1)(n-2).....(n-r+1)}{r!}{{(-x)}^{r}}+.......\]

(iii) \[{{(1-x)}^{-n}}=1+nx+\frac{n(n+1)}{2!}{{x}^{2}}+\frac{n(n+1)\,(n+2)}{3!}{{x}^{3}}+\]\[.....+\frac{n(n+1)......(n+r-1)}{r\,!}{{x}^{r}}+.....\]

(iv) \[{{(1+x)}^{-n}}=1-nx+\frac{n(n+1)}{2!}{{x}^{2}}-\frac{n(n+1)(n+2)}{3\,!}{{x}^{3}}+\]\[.....+\frac{n(n+1)......(n+r-1)}{r!}{{(-x)}^{r}}+......\]

(v) \[{{(1+x)}^{-1}}=1-x+{{x}^{2}}-{{x}^{3}}+.......\infty \]

(vi) \[{{(1-x)}^{-1}}=1+x+{{x}^{2}}+{{x}^{3}}+.......\infty \]

(vii) \[{{(1+x)}^{-2}}=1-2x+3{{x}^{2}}-4{{x}^{3}}+.......\infty \]

(viii) \[{{(1-x)}^{-2}}=1+2x+3{{x}^{2}}+4{{x}^{3}}+.......\infty \]

(ix) \[{{(1+x)}^{-3}}=1-3x+6{{x}^{2}}-...............\infty \]

(x) \[{{(1-x)}^{-3}}=1+3x+6{{x}^{2}}+.................\infty \]

Problems on approximation by the binomial theorem : We have \[{{(1+x)}^{n}}=1+nx+\frac{n(n-1)}{2!}{{x}^{2}}+.......\].

If \[x\] is small compared with 1, we find that the values of \[{{x}^{2}},{{x}^{3}},{{x}^{4}},.......\] become smaller and smaller.

\[\therefore \] The terms in the above expansion become smaller and smaller. If \[x\] is very small compared with 1, we might take 1 as a first approximation to the value of \[{{(1+x)}^{n}}\] or \[1+nx\] as a second approximation.