$3 + 13 + 29 + 51 + 79 + . . .$ to $n$ terms $=$

2 n^{2} + 7 n^{3}

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n^{2} + 5 n^{3}

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n^{3} + 2 n^{2}

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none of these

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Solution

The correct option is A $n^{3} + 2 n^{2}$
Let $S_{n} = 3 + 13 + 29 + 51 + 79 + \dots + n t e r m s$
$∵ S_{1} = 3$ and $S_{2} = 3 + 13 = 16$
Similarly $S_{3} = 3 + 13 + 29 = 45$
By options, from $(c)$ as we have $S_{n} = n^{3} + 2 n^{2}$
check by putting $n = 1, S_{1} = 1^{3} + {2.1}^{2} = 3$
$S_{2} = 2^{3} + {2.2}^{2} = 16$ and $S_{3} = 3^{3} + {2.3}^{2} = 45$
Thus option $(c)$ is correct

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