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Find the sum to n terms of the series 1/(1+1^2+1^4)+ 2/(1+2^2+2^4) + 3/(1+3^2+3^4) +.........

Find the sum to n terms of the series 1/(1+1^2+1^4)+ 2/(1+2^2+2^4) + 3/(1+3^2+3^4) +.........

Grade:10

1 Answers

mycroft holmes
272 Points
8 years ago
The general term is tn = n/(1+n2+n4) = ½ [1/(n2-n+1) – 1/(n2+n+1)]
 
Also n2+n+1 = (n+1)2-(n+1)+1.
 
So, the summation may be written as
 
½ [1/12-1+1) – 1/(22-2+1)] + ½ [1/(22-2+1) – 1/(32-3+1)]+...+½ [1/(n2-n+1) – 1/(n2+n+1)]
 
which simplifies to
½ [1 – 1/(n2+n+1)] 

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