What is the sum of n terms of the series 12 + | vedclass

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(D) Let the series be $S_n = 12 + 16 + 24 + 40 + \dots + T_n$.The differences between consecutive terms are $4, 8, 16, \dots$,which form a geometric p

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$4(2^n - 1) + 8n$

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(D) Let the series be $S_n = 12 + 16 + 24 + 40 + \dots + T_n$.The differences between consecutive terms are $4, 8, 16, \dots$,which form a geometric progression.The $n$-th term $T_n$ can be found using the method of differences:$T_n = T_1 + \sum_{k=1}^{n-1} (T_{k+1} - T_k) = 12 + \sum_{k=1}^{n-1} 2^{k+1} = 12 + (4 + 8 + 16 + \dots + 2^n)$.Using the sum of a geometric series formula,$T_n = 12 + \frac{4(2^{n-1} - 1)}{2 - 1} = 12 + 2^{n+1} - 4 = 2^{n+1} + 8$.Now,the sum of $n$ terms $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (2^{k+1} + 8)$.$S_n = (2^2 + 2^3 + \dots + 2^{n+1}) + (8 + 8 + \dots + 8)$.$S_n = \frac{4(2^n - 1)}{2 - 1} + 8n = 4(2^n - 1) + 8n$.

What is the sum of $n$ terms of the series $12 + 16 + 24 + 40 + \dots$?

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