find the sum to n terms of series:
1 + (1+ 1/2 ) + (1+ 1/2 + 1/4)...........to n terms.
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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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