The $n^{t h}$ term of the series $4 + 14 + 30 + 52 + 80 + 114 + \dots$ is

5 n - 1

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2 n^{2} + 2 n

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3 n^{2} + n

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2 n^{2} + 2

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Solution

The correct option is B $3 n^{2} + n$
$S = 4 + 14 + 30 + 52 + 80.... t (n)$
$S = 4 + 14 + 30 + 52..... t (n - 1) + t (n)$
Subtracting second from first
$t (n) = 4 + 10 + 16 + 22 + 28 + . . . . (4 + 6 (r - 1)) t (n) = 6 \times \frac{n (n + 1)}{2} - 2 n = 3 n^{2} + n$

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