Question

Let $I_n=\underset{0}{\overset{1}{\int }}{x}^n{\tan }^{-1}xdx.$ If $a_n{I}_{n+2}+b_n{I}_n=c_n$ for all $n\ge 1$, then

• A. $a_1,a_2,a_3$ are in G.P.
• B. $b_1,b_2,b_3$ are in A.P.
• C. $c_1,c_2,c_3$ are in H.P.
• D. $a_1,a_2,a_3$ are in A.P.

Answer 1

Answer: (B), (D)

The correct options are
B $b_1,b_2,b_3$ are in A.P.
D $a_1,a_2,a_3$ are in A.P.

$I_n=\underset{0}{\overset{1}{\int }}{x}^n{\tan }^{-1}xdx$
$I_n={\left[\frac{x^{n+1}}{n+1}{\tan }^{-1}x\right]}_0^1-\underset{0}{\overset{1}{\int }}\frac{x^{n+1}}{n+1}\left(\frac{1}{1+x^2}\right)dx$

$\Rightarrow (n+1)I_n=\frac{\pi }{4}-\underset{0}{\overset{1}{\int }}\frac{x^{n+1}}{1+x^2}dx\Rightarrow (n+3)I_{n+2}=\frac{\pi }{4}-\underset{0}{\overset{1}{\int }}\frac{x^{n+3}}{1+x^2}dx\Rightarrow (n+1)I_n+(n+3)I_{n+2}=\frac{\pi }{2}-\frac{1}{n+2}$

$\because a_n{I}_{n+2}+b_n{I}_n=c_n$
$\Rightarrow a_n=n+3,b_n=n+1,c_n=\frac{\pi }{2}-\frac{1}{n+2}$