Question

Find the value of
Sum =
$\frac{1}{3^{2} +1 } + \frac{1}{4^{2}+1} +\frac{1}{5^{2} +1} ...........\infty$ is equal to​

Answer 1

$\huge\bigcirc{\underline{\tt{\red{Answer}}}}\bigcirc$

$\huge{\boxed{\dfrac{13}{36}}}$

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Given:

$\large\sum\limits_{r=1}^{n} {\dfrac{1}{3^2 +1} + \dfrac{1}{4^2+1} + \dfrac{1}{5^2+1} .........\infty}$

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SoluTion:

General form

$\implies T_r= \dfrac{1}{(r+2)^2 + r } \\ \implies \dfrac{1}{r^2 + 5r +4 } \\ \implies \dfrac{1}{(r+1)(r+4)}$

1 can be written as (r+4)-(r+1)

$\leadsto \dfrac{(r+4)-(r+1)}{(r+1)(r+4)} \\ \implies \bigg(\dfrac{\cancel{(r+4)}}{\cancel{(r+4)}(r+1)} - \dfrac{\cancel{(r+1)}}{\cancel{(r+1)}(r+4)} \bigg) \\ \implies \bigg( \dfrac{1}{r+1} - \dfrac{1}{r+4} \bigg)$

Putting Values from 1 to n in r

$\dfrac{1}{3} \bigg[ \bigg(\dfrac{1}{2}-\cancel{\dfrac{1}{5}}\bigg) +\bigg( \dfrac{1}{3} -\cancel{\dfrac{1}{6}}\bigg) +\bigg(\dfrac{1}{4} -\cancel{\dfrac{1}{7}}\bigg) + \bigg(\cancel{\dfrac{1}{5}} - \cancel{\dfrac{1}{8}}\bigg) .......rest\: all\:gets\:cancelled \bigg] \\ \implies \dfrac{1}{3} \bigg( \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} \bigg) \\ \implies \dfrac{1}{3} \bigg( \dfrac{12 +8 +6}{24 } \bigg) \\ \leadsto \dfrac{\cancel{26}}{\cancel{72}} \\ \boxed{\dfrac{13}{36}}$