$1^{3} + 1^{2} + 1 + 2^{3} + 2^{2} + 2 + 3^{3} + 3^{2} + 3 + . . . 3 n$ terms $=$

\frac{n (n + 1) (n^{2} + 12 n + 5)}{12}

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\frac{n (n + 1) (3 n^{2} + 7 n + 8)}{12}

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\frac{n (n + 1) (n + 2) (n^{2} + 5 n + 6)}{12}

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\frac{(n + 1) (n + 2) (n + 3)}{4}

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Solution

The correct option is B $\frac{n (n + 1) (3 n^{2} + 7 n + 8)}{12}$
$S = Σ r^{3} + Σ r^{2} + Σ r$

$S = \frac{n^{2} (n + 1)^{2}}{4} + \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}$

$S = \frac{n (n + 1) (3 n^{2} + 7 n + 8)}{12}$

Hence Option B

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Similar questions

The sum of the series $1^{2}$ .2 + $2^{2}$ .3 + $3^{2}$ .4 + ........ to n terms is

6

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

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1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + \dots

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= \frac{n {(n + 1)}^{2} (n + 2)}{12}

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