Sequences & Series
Telescoping series with partial fractions
Grade Class 12
Question:
The value of $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$ to $\infty$ is
$\dfrac{31}{13}$
$\dfrac{13}{36}$
$\dfrac{31}{36}$
$\dfrac{1}{36}$
Step-by-Step Solution
Key Concept: General term: $\dfrac{1}{n^2+(n-2)}=\dfrac{1}{n^2+n-2}=\dfrac{1}{(n+2)(n-1)}=\dfrac{1}{3}\left(\dfrac{1}{n-1}-\dfrac{1}{n+2}\right)$. Starting $n=3$: telescope.
$\dfrac{13}{36}$.
Correct Answer: 2