Sum of n terms of series 12 + 16 + 24 + 40 +......... will be

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Solution

The correct option is D

Let n terms of series is $T_{n}$ then

$S_{n}$ = 12 + 16 + 24 + 40 + ........ + $T_{n}$

Again $S_{n}$ = 12 + 16 + 24 +.........+ $T_{n}$

On Subtraction

0 = (12 + 4 + 8 + 16 +........+ upto n terms) - $T_{n}$

or $T_{n}$ = 12 + [ 4 + 8 + 16 +........ + upto (n-1) terms]

= 12 + $\frac{4 (2^{n - 1} - 1)}{2 - 1}$ = $2^{n + 1}$ + 8

On putting n = 1, 2, 3........

$T_{1}$ = $2^{2}$ + 8, $T_{2}$ = $2^{3}$ + 8, $T_{3}$ = $2^{4}$ + 8......... etc.

$S_{n}$ = $T_{1}$ + $T_{2}$ + $T_{3}$ + .......+ $T_{n}$

= ( $2^{2} + 2^{3} + 2^{4} + . . . . . .$ upto n terms) + (8 + 8 + 8 +........upto n terms)

= $\frac{2^{2} (2^{n} - 1)}{2 - 1}$ + 8 n = 4 ( $2^{n}$ - 1) + 8n.

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