Let Sn denote the sum of the cubes of the first n natural numbers and Sn denote the sum of the first natural numbers. Then ∑r=1n SrSr equals(a) nn+1n+26(b) nn+12(c) n2+3n+22(d) None of these

Let Sn denote the sum of the cubes of the first n natural numbers and

1.	Let Sn denote the sum of the cubes of the first n natural numbers and Sn denote the sum of the first natural numbers. Then ∑r=1n SrSr equals(a) nn+1n+26(b) nn+12(c) n2+3n+22(d) None of these
Answer» Let S_n denote the sum of the cubes of the first n natural numbers and S_n denote the sum of the first natural numbers. Then $\sum_{r = 1}^{n} \frac{S_{r}}{S_{r}}$ equals (a) $\frac{n (n + 1) (n + 2)}{6}$ (b) $\frac{n (n + 1)}{2}$ (c) $\frac{n^{2} + 3 n + 2}{2}$ (d) None of these