Question

If $C_0,C_1,C_2,C_3,....$ are binomial coefficients in the expansion of $(1+x{)}^n$ , then $\frac{C_0}{3}-\frac{C_1}{4}+\frac{}{}-\frac{}{}+...$ is equal to :

• A. $\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}$
• B. $\frac{1}{n+1}+\frac{2}{n+2}-\frac{3}{n+3}$
• C. $\frac{1}{n+2}-\frac{1}{n+1}+\frac{1}{n+3}$
• D. $\frac{2}{n+1}-\frac{1}{n+2}+\frac{2}{n+3}$
• E. $\frac{1}{n+2}-\frac{2}{n+1}+\frac{3}{n+3}$

Answer 1

The correct option is A $\frac{1}{n+1}-\frac{2}{n+2}+\frac{1}{n+3}$

We know $(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-...+(-1{)}^n{C}_n\cdot x^n$

On multiplying both sides bby $x^2$ , we get

$(1-x{)}^n{x}^2=C_0{x}^2-C_1{x}^3+C_2{x}^4-...+$

On integrating both sides by taking limit $0$ to $1$.

Therefore, ${\int }_0^1(1-x{)}^n{x}^{2}dx={\int }_0^1(C_0{x}^2-C_1{x}^3+C_2{x}^4+....)dx$

${\int }_0^1{x}^n(1-x{)}^{2}dx={[C_0\frac{x^3}{3}-C_1\frac{x^4}{4}+C_2\frac{x^5}{5}-...]}_0^1$

$\Rightarrow {\int }_0^1{x}^n(1+x^2-2x)dx=\frac{C_0}{3}-\frac{C_1}{4}+\frac{C_2}{5}-...$

Here $\frac{C_0}{3}-\frac{C_1}{4}+\frac{C_2}{5}-...$

$={[\frac{x^{n+1}}{n+1}+\frac{x^{n+3}}{n+3}-\frac{2x^{n+2}}{n+2}]}_0^1$

$=[\frac{1}{n+1}+\frac{1}{n+3}-\frac{2}{n+2}]$