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  • Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then sumr=1tonS_rdivideS_r equals equals. A.

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} \frac{S_{r}}{s_{r}} equals.
A. \frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{6} \\

B. \frac{\mathrm{n}(\mathrm{n}+1)}{2} \\
C.  \frac{\mathrm{n}^{2}+3 \mathrm{n}+2}{2}
D. None of these

Answers (1)

Sum of cubes of first n natural numbers  S_{n}= \sum _{i=1}^{n}i^{3}= \left( \frac{n \left( n+1 \right) }{2} \right) ^{2} \\\\

 

Sum of first n natural numbers    s_{n}= \sum _{i=1}^{n}i=\frac{n \left( n+1 \right) }{2} \\\\


    \\ \sum _{i=1}^{n}\frac{S_{n}}{s_{n}}= \sum _{i=1}^{n}\frac{n^{2}}{2}+\frac{n}{2} \\\\ =\frac{1}{2} \left[ \sum _{i=1}^{n}n^{2}+ \sum _{i=1}^{n}n \right] \\\\ =\frac{1}{2} \left[ \frac{n \left( n+1 \right) \left( 2n+1 \right) }{6}+\frac{n \left( n+1 \right) }{2} \right] \\\\ =\frac{1}{2} \left[ \frac{n \left( n+1 \right) }{2} \left( \frac{2n+1}{3}+1 \right) \right] \\\\
\\ =\frac{1}{2}\frac{n \left( n+1 \right) }{2}\frac{ \left( 2n+4 \right) }{3} \\\\ =\frac{n \left( n+1 \right) \left( n+2 \right) }{6} \\\\

 

Hence, correct option is (a).

Posted by

infoexpert21

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