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Question

$$\displaystyle 1^{2}+\left ( 1^{2}+2^{2} \right )+\left (1^{2}+2^{2}+3^{2} \right )+...$$ to $$n$$ terms.


A
112n(n+1)(2n3+3n2+n)
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B
112n(n+1)(3n3+2n2+n)
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C
16n(n+1)(2n+1)
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D
14n2(n+1)2
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Solution

The correct option is A $$\dfrac{1}{12}n(n+1)(2n^3+3n^2+n)$$
nth term $$T_n = 1^2+2^2+3^2+.....+n^2=\sum n^2=\cfrac{1}{6}n(n+1)(2n+1)=\cfrac{1}{6}(2n^3+3n^2+n)$$
Thus required summation is $$\displaystyle =\sum T_n =\frac{1}{6}\left(\sum 2n^3+\sum 3n^2+ \sum n\right) $$
$$\displaystyle = \frac{1}{6}\left(2\frac{n^2(n+1)^2}{4}+3\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right) =\frac{1}{12}n(n+1)(2n^3+3n^2+n)$$

Mathematics

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