Question

(C₀ / 1) + (C₂ / 3) + (C₄ / 5) + (C₆ / 7) + ….. = +

• A. $\frac{2^{n+1}}{(n+1)}$
• B. $\frac{2^{n+1}-1}{n+1}$
• C. $\frac{2^n}{(n+1)}$
• D. None of these

Answer 1

The correct option is C

$\frac{2^n}{(n+1)}$

Explanation for correct options:

Step 1: Declaring the theorem

We know, ${(1+x)}^n={}^n{C}_0{x}^0+{}^n{C}_1{x}^1+{}^n{C}_2{x}^2+{}^n{C}_3{x}^3....+{}^n{C}_n{x}^n$ .. (1)

${(1-x)}^n={}^n{C}_0{x}^0-{}^n{C}_1{x}^1-{}^n{C}_2{x}^2-{}^n{C}_3{x}^3....{(-1)}^n{}^n{C}_n{x}^n$ …(2)

Step 2: Integrating the equations

Equation 1: ${\int }_0^1{(1+x)}^n={}^n{C}_0{\int }_0^{1}d{x}^{}+{}^n{C}_1{\int }_0^{1}d{x}^{}+....+{}^n{C}_n{\int }_0^1{x}^{n}dx$

$\Rightarrow {\frac{(1+x)}{n+1}}^{n+1}={}^n{C}_0+\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{3}+....+\frac{{}^n{C}_n}{n+1}$… (3)

Equation 2: ${\int }_0^1{(1-x)}^n={}^n{C}_0{\int }_0^{1}d{x}^{}+{}^n{C}_1{\int }_0^{1}d{x}^{}+....+(-1){{\int }_0^1}^{}{C}_n{x}^{n}dx$

$\Rightarrow {\frac{(1-x)}{n+1}}^{n+1}={}^n{C}_0-\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{3}+....+(-1)\frac{{}^n{C}_n}{n+1}$ … (4)

Step 3: Evaluating the equations

Equation (3) + Equation (4)

$\Rightarrow {\frac{(1+x)}{n+1}}^{n+1}-{\frac{(1-x)}{n+1}}^{n+1}=2({}^n{C}_0+\frac{{}^n{C}_2}{3}+\frac{{}^n{C}_4}{5}...)$

$\Rightarrow {\frac{(1+x)}{n+1}}^{n+1}-{\frac{(1-x)}{n+1}}^{n+1}=2({}^n{C}_0+\frac{{}^n{C}_2}{3}+\frac{{}^n{C}_4}{5}...)$

Applying the limits:

$\Rightarrow \frac{2^{n+1}-1}{n+1}+\frac{1}{n+1}=2({}^n{C}_0+\frac{{}^n{C}_2}{3}+\frac{{}^n{C}_4}{5}...)$

$\Rightarrow \frac{2^{n+1}}{n+1}=2({}^n{C}_0+\frac{{}^n{C}_2}{3}+\frac{{}^n{C}_4}{5}...)$

$\Rightarrow \frac{2^n}{n+1}=({}^n{C}_0+\frac{{}^n{C}_2}{3}+\frac{{}^n{C}_4}{5}....)$

Therefore, the expression $({}^n{C}_0+\frac{{}^n{C}_2}{3}+\frac{{}^n{C}_4}{5}....)$= $\frac{2^n}{n+1}$

Hence, Option (C) is the correct .