Sum of n terms of series $16 + 24 + 40 + . . . . . . . .$ will be.

4 (2^{n} - 1) + 8 n

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6 (2^{n} - 1) + 6 n

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3 (2^{n} - 1) + 8 n

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8 (2^{n} - 1) + 8 n

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Solution

The correct option is A $8 (2^{n} - 1) + 8 n$
$S_{n} = 16 + 24 + 40 + \dots + T_{n}$
$= (8 + 2^{3}) + (8 + 2^{4}) + \dots + (8 + 2^{n + 2})$
$= 8 n + (2^{3} + 2^{4} + \dots + 2^{n + 2}) = 8 n + [\frac{2^{3}}{2 - 1} (2^{n} - 1)] = 8 n + 8 (2^{n} - 1)$

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