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  4. If λ= displaystyle∑n=1∞ (1/9 n2+3 n-2), then (1/λ) equals

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Q. If $\lambda=\displaystyle\sum_{n=1}^{\infty} \frac{1}{9 n^2+3 n-2}$, then $\frac{1}{\lambda}$ equals

Sequences and Series

Solution:

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