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` equals ______.

#### Options

`(n(n + 1)(n + 2))/6`

`(n(n + 1))/2`

`(n^2 + 3n + 2)/2`

None of these

#### Solution

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` equals **`(n(n + 1)(n + 2))/6`**.

**Explanation:**

Given that `sum_(i = 1)^n S_r/s_r = S_1/s_1 + S_2/s_2 + S_3/s_3 + ... + S_n/s_n`

Let T_{n} be the nth term of the above series

∴ T_{n} = `S_n/s_n`

= `([(n(n + 1))/2]^2)/((n(n + 1))/2)`

= `(n(n + 1))/2`

= `(n^2 + n)/2`

Now sum of the given series

`sum"T"_"n" = 1/2 sum [n^2 + n]`

= `1/2 [sum n^2 + sum n]`

= `1/2 [(n(n + 1)(2n + 1))/6 + (n(n + 1))/2]`

= `1/2 * (n(n + 1))/2 [(2n + 1)/3 + 1]`

= `(n(n + 1))/4 [(2n + 1 + 3)/3]`

= `(n(n + 1))/4 * ((2n + 4))/3`

= `(n(n + 1)(n + 2))/6`