JEE Main 2017 · 9 Apr · Shift 1 · Mathematics Q69

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Let $S_n=\dfrac1{1^3}+\dfrac{1+2}{1^3+2^3}+\dfrac{1+2+3}{1^3+2^3+3^3}+\dots+\dfrac{1+2+\dots+n}{1^3+2^3+\dots+n^3}$ . If $100S_n=n$ , then n is equal to :

199

200

Answer and worked solution

Official answer: A

The $k$ -th term is $\dfrac{k(k+1)/2}{k^2(k+1)^2/4}=\dfrac2{k(k+1)}$ , so $S_n=2\left(1-\frac1{n+1}\right)=\dfrac{2n}{n+1}$ . Then $\dfrac{200n}{n+1}=n$ gives $n=199$ .

Topics:Special series sums

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JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 69