The sum of the following series 1+6+912+22+327+1212+22+32+429+1512+22+…+5211+… upto 15 terms, is

# The sum of the following 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}+\dots +{5}^{2}\right)}{11}+\dots$ upto 15 terms, is

1. A

7510

2. B

7820

3. C

7830

4. D

7520

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### Solution:

The  term of given expression is

${T}_{n}=\frac{\left(3+\left(n-1\right)3\right)\left({1}^{2}+{2}^{2}+{3}^{2}+\dots +{n}^{2}\right)}{\left(2n+1\right)}$

$=\frac{3n\cdot \frac{n\left(n+1\right)\left(2n+1\right)}{6}}{2n+1}=\frac{{n}^{2}\left(n+1\right)}{2}$

Now, ${S}_{15}=\frac{1}{2}\sum _{n=1}^{15} \left({n}^{3}+{n}^{2}\right)$ $=\frac{1}{2}\left[{\left(\frac{15\left(15+1\right)}{2}\right)}^{2}+\frac{15×16×31}{6}\right]$

$=\frac{1}{2}\left[\left(120{\right)}^{2}+1240\right]=7820$

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