Question

The sum of the follwing series $1 + 6 + \frac{9\left(1^{2} + 2^{2} +3^{2}\right)}{7} + \frac{12\left(1^{2} + 2^{2} +3^{2} +4^{2}\right)}{9} + \frac{15\left(1^{2} +2^{2} +.... + 5^{2}\right)}{11} + .....$ up to $15$ terms, is:

• A. 7820
• B. 7830
• C. 7520
• D. 7510

Answer 1

Answer: (A)

$T_{n} = \frac{\left(3+ \left(n-1\right) \times 3\right)\left(1^{2} +2^{2} +.... +n^{2}\right)}{\left(2n+1\right)} $
$ T_{n} = \frac{3. \frac{n\left(n+1\right)\left(2n+1\right)}{6}}{2n+1}= \frac{n^{2} \left(n+1\right)}{2} $
$ S_{15} = \frac{1}{2} \sum^{15}_{n=1} \left(n^{3} +n^{2}\right) = \frac{1}{2} \left[\left(\frac{15 \left(15 +1\right)}{2}\right)^{2} + \frac{15 \times 16 \times 31}{6}\right]$
$ = 7820 $