Question

Find the value of $\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+.....$

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

Answer 1

Answer: (D)

None of the above

Consider the expansion ${(1-x)}^n=C_0-C_{1}x+C_2{x}^2-...{(-1)}^n{x}^n$,

Multiplying with $x$ on both sides gives,

$x{(1-x)}^n=C_{0}x-C_1{x}^2+C_2{x}^3-...{(-1)}^n{x}^{n+1}$,

Integrating with respect to $x$ from $0$ to $1$ gives,

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

From the first principle, i.e.,${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

${\int }_0^1(1-x)x^{n}dx⟹{\int }_0^1{x}^n-x^{n+1}dx⟹\frac{1}{n+1}-\frac{1}{n+2}⟹\frac{1}{(n+1)(n+2)}$