Question

If $\lambda =\sum _{n=1}^{\infty }\frac{1}{9n^2+3n-2}$, then $\frac{1}{\lambda }$ equals

• A. 4
• B. 6
• C. 10
• D. 20

Answer 1

Answer: (B)

$\lambda =\sum _{n=1}^{\infty }\frac{1}{(3n+2)(3n-1)}$
$T_n=\frac{1}{3}\left[\frac{(3n+2)-(3n-1)}{(3n+2)(3n-1)}\right]$
$=\frac{1}{3}\left[\frac{1}{3n-1}-\frac{1}{3n+2}\right]$
$\Rightarrow \lambda =\frac{1}{6}\Rightarrow \frac{1}{\lambda }=6$