Let `S_n` denote the sum of the cubes of the first `n` natural numbers and `s_n` denote the sum of the first `n` natural numbers. Then `sum_(r=1)^n S_r/s_r` is equal to

Explanation

We have given `S_n` denote the sum of cubes of the first n natural numbers.

Hence `S_n=(n^2(n+1)^2)/4`

And, `s_n` denote the sum of the first n natural numbers.
Hence `s_n=(n(n+1))/2`

Now,

`\Rightarrow T_n=\sum_{r=1}^{n}(S_r/s_r)=\sum_{r=1}^{n}(r(r+1))/2`

`\Rightarrow T_n=(1/2)(\sum_{r=1}^{n}r^2+\sum_{r=1}^{n}r)=((n(n+1)(2n+1))/12)+((n(n+1))/4)`

`\Rightarrow T_n=(n(n+1)(n+2))/6`

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