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# Sum of series $S$ = $\frac{1}{2}$+$\frac{1}{3}\;(1+2)$+$\frac{1}{4}\;(1+2+3)$+$\frac{1}{5}\;(1+2+3+4)$+......upto 20 terms is

$(a)\;110\qquad(b)\;105\qquad(c)\;115\qquad(d)\;116$

Can you answer this question?

Answer : (b) 105
Explanation : $k^{th}\;term\;t_{k}=\frac{1}{k+1}\;[1+2+\;.....\;+k]$
$=\frac{1}{k+1}\;\frac{k(k+1)}{2}\;=\frac{k}{2}$
$S=\sum_{k=1}^{20}\;\frac{k}{2}=\frac{1}{2}*\frac{20*21}{2}=105.$
answered Jan 20, 2014 by