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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

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detailed solution

Correct option is C

Let tk denote the kth term of the series. Thentk=1k(1+2+3+…+k)=1kk(k+1)2=k+12.Thus, S=12[2+3+4+…+21]=102(2+21)=115

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

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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