Question

$\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}......=$

Answer 1

We know that

${(1+x)}^n=C_0^n+C_1^{n}x+C_2^n{x}^2+C_3^n{x}^3.......+C_n^n{x}^n-\left(1\right){(1-x)}^n=C_0^n-C_1^{n}x+C_2^n{x}^2-C_3^n{x}^3.......+{(-1)}^n{C}_n^n{x}^n-\left(2\right)$

Integrating both sides of eq$\left(1\right)$ under limits $0$ to $1$ we get

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

Integrating both sides of eq$\left(2\right)$ under limits $0$ to $1$ we get

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

Adding $\left(3\right)$ & $\left(4\right)$ we get

${\Rightarrow \frac{{(1+x)}^{n+1}}{n+1}}_0^1-{\frac{{(1-x)}^{n+1}}{n+1}}_0^1=2(C_0^n+\frac{C_2^n}{3}+\frac{C_4^n}{5}.......)\Rightarrow \frac{2^{n+1}-1}{n+1}+\frac{1}{n+1}=2(C_0^n+\frac{C_2^n}{3}+\frac{C_4^n}{5}.....)\Rightarrow \frac{2^{n+1}}{n+1}=2(C_0^n+\frac{C_2^n}{3}+\frac{C_4^n}{5}.....)\therefore \frac{C_0^n}{1}+\frac{C_2^n}{3}+\frac{C_4^n}{5}.......=\frac{2^n}{n+1}$