Question

Let $I_{n} = \int^{1}_{0} x^{n} tan^{-1} x\,dx.$ 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)

$\left(n+1\right)I_{n} + \left(n+3\right)I_{n+2} = \int^{1}_{0} \left\{\left(n+1\right)x^{n}+\left(n+3\right)x^{n+2}\right\}tan^{-1}x\,dx$
$=\left[\left\{\left(n+1\right) \frac{x^{n+1}}{n+1}+\left(n+3\right) \frac{x^{n+3}}{n+3}\right\}tan^{-1} x\right]^{1}_{0} -\int^{1}_{0} x^{n+1}\left(1+x^{2}\right). \frac{1}{1+x^{2}}dx = \left[\left(x^{n+1} +x^{n+3}\right)tan^{-1} x\right]^{1}_{0}-\left[\frac{x^{n+2}}{n+2}\right]^{1}_{0}$
$= \frac{\pi}{2} - \frac{1}{n+2}$
$c_{n} = \frac{\pi}{2}-\frac{1}{n+2}$
$\therefore \, a_{n} = n+3; b_{n} = n+1; c_{n} = \frac{\pi}{2} - \frac{1}{n+2}$