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Mathematics · 2017

JEE Main · 9 April 2017, Shift 1 · Q69

Let S_n =1/1^3+(1+2)/(1^3+2^3)+(1+2+3)/(1^3+2^3+3^3)+ … …+(1+2+… …..+ n)/(1^3+2^3+… ….+ n^3). If 100 S_n = n then n is equal to:

Let $$\begin{aligned} & \mathrm{S}_{\mathrm{n}}=\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+ \\ & \ldots \ldots+\frac{1+2+\ldots \ldots . .+\mathrm{n}}{1^3+2^3+\ldots \ldots .+\mathrm{n}^3} \text {. If } 100 \mathrm{~S}_{\mathrm{n}}=\mathrm{n} \\ & \text { then } \mathrm{n} \text { is equal to : } \end{aligned} $$
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JEE Main 2017 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.