Question

Let $I_n={\int }_0^1{x}^{n}ta{n}^{-1}xdx.$ 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)

$\left(n+1\right)I_n+\left(n+3\right)I_{n+2}={\int }_0^1\left\{\left(n+1\right)x^n+\left(n+3\right)x^{n+2}\right\}ta{n}^{-1}xdx$
$={\left[\left\{\left(n+1\right)\frac{x^{n+1}}{n+1}+\left(n+3\right)\frac{x^{n+3}}{n+3}\right\}ta{n}^{-1}x\right]}_0^1-{\int }_0^1{x}^{n+1}\left(1+x^2\right).\frac{1}{1+x^2}dx={\left[\left(x^{n+1}+x^{n+3}\right)ta{n}^{-1}x\right]}_0^1-{\left[\frac{x^{n+2}}{n+2}\right]}_0^1$
$=\frac{\pi }{2}-\frac{1}{n+2}$
$c_n=\frac{\pi }{2}-\frac{1}{n+2}$
$\therefore a_n=n+3;b_n=n+1;c_n=\frac{\pi }{2}-\frac{1}{n+2}$