Question

$\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+.........+\frac{C_{16}}{17}=$

• A. $\frac{2^{15}}{14}$
• B. $\frac{2^{16}}{17}$
• C. $\frac{2^{15}}{16}$
• D. $\frac{2^{20}}{22}$

Answer 1

Answer: (B)

$\frac{2^{16}}{17}$

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

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

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

Substituting $n=16$ gives,

${\int }_0^1{(1+x)}^{n}dx=C_0+\frac{C_1}{2}+\frac{C_2}{3}+...+\frac{C_n}{n+1}⟹C_0+\frac{C_1}{2}+\frac{C_2}{3}+...+\frac{C_{16}}{17}=\frac{2^{17}-1}{17}$

Similarly,integrating with respect to $x$ from $0$ to $-1$ gives,

$\frac{-1}{17}=C_0-\frac{C_1}{2}+\frac{C_2}{3}-...+\frac{C_{16}}{17}$

adding these two equations gives,

$C_0+\frac{C_2}{3}+...+\frac{C_{16}}{17}=\frac{2^{16}}{17}$