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

5n - 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 $3 n^{2} + n$
Let $n^{t h}$ term of the series is $T_{n}$ and
$S_{n} = 4 + 14 + 30 + 52 + 80 + 114 + . . . . . + T_{n} . . (i)$
$S_{n} = 4 + 14 + 30 + 52 + 80 + . . . . . + T_{n} . . (i i)$
Subtract (ii) from (i)
0 = (4 + 10 + 16 + 22 + 28 + 34 + ..... n terms) - $T_{n}$
$T_{n}$ = 4 + 10 + 16 + 22 + ... n terms
= $\frac{n}{2} [2 \times 4 + (n - 1) 6] = n (3 n + 1) = 3 n^{2} + n$

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