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Question: If t<sub>n</sub> = n(n + 1) (n + 2) then \(\sum _ { n = 1 } ^ { \infty } \frac { 1 } { t _ { n } }\...

If tn = n(n + 1) (n + 2) then ∑n=1∞1tn\sum _ { n = 1 } ^ { \infty } \frac { 1 } { t _ { n } } is equal to

A

½

B

¼

C

1

D

None of these

Answer

¼

Explanation

Solution

1tn\frac { 1 } { \mathrm { t } _ { \mathrm { n } } } = 1n(n+1)(n+2)\frac { 1 } { \mathrm { n } ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) } =12\frac { 1 } { 2 } [1n(n+1)−1(n+1)(n+2)]\left[ \frac { 1 } { n ( n + 1 ) } - \frac { 1 } { ( n + 1 ) ( n + 2 ) } \right]

\S= ∑n=1n1tn\sum _ { n = 1 } ^ { n } \frac { 1 } { t _ { n } } =12\frac { 1 } { 2 }×[(11.2−12.3)+(12.3−13.4)+…\left[ \left( \frac { 1 } { 1.2 } - \frac { 1 } { 2.3 } \right) + \left( \frac { 1 } { 2.3 } - \frac { 1 } { 3.4 } \right) + \ldots \right..

... +(1n(n+1)−1(n+1)(n+2))]\left. + \left( \frac { 1 } { \mathrm { n } ( \mathrm { n } + 1 ) } - \frac { 1 } { ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) } \right) \right]

Sn =14\frac { 1 } { 4 }