Sum of the series S=1+12(1+2)+13(1+2+3)+14(1+2+3+4)+…up to 20 terms is

Sum of the series $S = 1 + \frac{1}{2} (1 + 2) + \frac{1}{3} (1 + 2 + 3) + \frac{1}{4} (1 + 2 + 3 + 4) + \dots$ up to 20 terms is

Let $t_{k}$ denote the kth term of the series. Then

$t_{k} = \frac{1}{k} (1 + 2 + 3 + \dots + k) = \frac{1}{k} \frac{k (k + 1)}{2} = \frac{k + 1}{2} .$

Thus, $S = \frac{1}{2} [2 + 3 + 4 + \dots + 21] = \frac{10}{2} (2 + 21) = 115$

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