Question

$\frac{C_0}{1}-\frac{C_1}{4}+\frac{C_2}{7}-...+{(-1)}^n\frac{C_n}{3n+1}=$_____ where $C_r$ stands for ${{}^{n}C}_r$

• A. $\frac{3^n}{\mathrm{1.4.7...}(3n+1)}$
• B. $\frac{3.n!}{\mathrm{1.4.7...}(3n+1)}$
• C. $\frac{3^{n+1}.n!}{\mathrm{1.4.7...}(3n+1)}$
• D. $\frac{3^n.n!}{\mathrm{1.4.7...}(3n+1)}$

Answer 1

Answer: (D)

$\frac{3^n.n!}{\mathrm{1.4.7...}(3n+1)}$

${(1-x^3)}^n=C_0-C_1{x}^3+C_2{x}^6-C_3{x}^9+\cdots +(-1{)}^n{C}_n{x}^{3n}$
Integrating both sides with in limits $0$ to $1$,
${\int }_0^1{(1-x^3)}^{n}dx=\frac{C_0}{1}-\frac{C_1}{4}+\frac{C_2}{7}-...+{(-1)}^n\frac{C_n}{3n+1}$
let $I_n={\int }_0^1{(1-x^3)}^{n}dx$ integrating by parts
$I_n=\left[x{(1-x^3)}^n\right]\begin{array}{c}1\\ 0\end{array}-{\int }_0^1{x.n(1-x^3)}^{n-1}(-3x^2)dx$
$=0-3n{\int }_0^1{(1-x^3)}^{n-1}(1-x^3-1)dx$
$\Rightarrow I_n=-3n{I}_n+3n{I}_{n-1}$
$\to I_n(3n+1)=3n{I}_{n-1}$
$\therefore I_n=\frac{3n}{3n+1}{I}_{n-1}=\frac{3n}{3n+1}\cdot \frac{3n-3}{3n-2}\cdot \cdot \cdot \frac{6}{7}\cdot \frac{3}{4}{I}_0$
where $I_0={\int }_0^1{(1-x^3)}^{0}dx={\int }_0^{1}1dx=1$
$\Rightarrow I_n=\frac{3^n.n!}{\mathrm{1.4.7...}(3n+1)}$
$\therefore \frac{C_0}{1}-\frac{C_1}{4}+\frac{C_2}{7}-...+{(-1)}^n\frac{C_n}{3n+1}=\frac{3^n.n!}{\mathrm{1.4.7...}(3n+1)}$
Hence, option D.