Question

$C_0-\frac{C_1}{2}+\frac{C_2}{3}-.......+\frac{(-1{)}^n{C}_n}{n+1}$ is

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

Answer 1

The correct option is C $\frac{1}{n+1}$

$1-(1-x{)}^n=C_{1}x-C_2{x}^2+....+(-1{)}^{n-1}{C}_n{x}^n$
Integrating within the limits 0 to 1, we get
${\int }_0^1\frac{1-(1-x{)}^n}{x}dx=C_1-\frac{C_2}{2}+\frac{C_3}{3}+......+\frac{(-1{)}^{n-1}{C}_n}{n}$
Thus R.H.S.$={\int }_0^1\frac{1-x^n}{1-x}dx$
$={\int }_0^1(1+x+x^2+...+x^{n-1})dx$
$=1+\frac{1}{2}+........+\frac{1}{n}.$ for Example 72
${\int }_0^1(1-x{)}^{n}dx={\int }_0^1(1-C_{1}x+C_2{x}^2+.....+(-1{)}^n{C}_n{x}^n)dx$
L.H.S.$={\int }_0^1{x}^{n}dx=\frac{1}{n+1}$ and the R.H.S. is equal to
$C_0-\frac{C_1}{2}+\frac{C_2}{3}-.......+\frac{(-1{)}^n{C}_n}{n+1}$