Question

The value of $C_1-\frac{C_2}{2}+\frac{C_3}{3}+....+\frac{(-1{)}^{n-1}}{n}{C}_n$ is

• A. $\sum _{k=1}^n\frac{1}{k}$
• B. $\sum _{k=1}^n\frac{(-1{)}^{k+1}}{k}$
• C. $\sum _{k=1}^n\frac{1}{k^2}$
• D. $\sum _{k=1}^n\frac{(-1{)}^{k+1}}{k^2}$

Answer 1

Answer: (A)

$\sum _{k=1}^n\frac{1}{k}$

$I_n={\int }_0^1(1-x^2{)}^{n}dx$
$={\int }_0^1(1-C_1{x}^2+C_2{x}^4+......+(-1{)}^n{C}_n{x}^{2n})dx$
$=1-\frac{C_1}{3}+\frac{C_2}{5}-......-\frac{(-1{)}^n}{2n+1}$
But $I_n=\frac{\left(2n\right)!!}{(2n+1)!!}$
For Example 71, consider
$1-(1-x{)}^n=C_{1}x-C_2{x}^2+....+(-1{)}^{n-1}{C}_n{x}^n$
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
$C_0-\frac{C_1}{2}+\frac{C_2}{3}-.......+\frac{(-1{)}^n{C}_n}{n+1}$