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Question: The sum to n terms of the series \(\left[ \frac { 1 } { 1 + 1 ^ { 2 } + 1 ^ { 4 } } + \frac { 2 } ...

The sum to n terms of the series

[11+12+14+21+22+24+31+32+34+…..+n1+n2+n4]\left[ \frac { 1 } { 1 + 1 ^ { 2 } + 1 ^ { 4 } } + \frac { 2 } { 1 + 2 ^ { 2 } + 2 ^ { 4 } } + \frac { 3 } { 1 + 3 ^ { 2 } + 3 ^ { 4 } } + \ldots . . + \frac { n } { 1 + n ^ { 2 } + n ^ { 4 } } \right] is-

A

B

C

D

None of these

Answer

Explanation

Solution

Tn = = n(1+n2)2−n2\frac { \mathrm { n } } { \left( 1 + \mathrm { n } ^ { 2 } \right) ^ { 2 } - \mathrm { n } ^ { 2 } }

= n(1+n2−n)(1+n2+n)\frac { \mathrm { n } } { \left( 1 + \mathrm { n } ^ { 2 } - \mathrm { n } \right) \left( 1 + \mathrm { n } ^ { 2 } + \mathrm { n } \right) }

= 12\frac { 1 } { 2 }

Sn = 12\frac { 1 } { 2 }

̃ Sn = 12\frac { 1 } { 2 } = n(n+1)2(n2+n+1)\frac { \mathrm { n } ( \mathrm { n } + 1 ) } { 2 \left( \mathrm { n } ^ { 2 } + \mathrm { n } + 1 \right) }