$n^{t h}$ term of the series $4, 14, 30, 52, . . .$

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 C $3 n^{2} + n$
Let the nth term be $a n^{2} + b n + c$ , then
First term $= a (1)^{2} + b (1) + c = a + b + c = 4$
Second term $= a (2)^{2} + b (2) + c = 4 a + 2 b + c = 14$
Third term $= a (3)^{2} + b (3) + c = 9 a + 3 b + c = 30$
Solving these three equations, we get
$a = 3, b = 1, c = 0$
Therefore,
nth term $= 3 n^{2} + n$

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