What is the sum of n terms of the series 2 + | vedclass

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(B) The given series is $2, 5, 14, 41, \dots$The differences between consecutive terms are $3, 9, 27, \dots$,which form a geometric progression.The $n

Vedclass · Accepted Answer

$\frac{n}{2} + \frac{3}{4}(3^n - 1)$

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(B) The given series is $2, 5, 14, 41, \dots$The differences between consecutive terms are $3, 9, 27, \dots$,which form a geometric progression.The $n$-th term $t_n$ can be written as:$t_n = 2 + (3 + 9 + 27 + \dots + 3^{n-1})$Using the sum formula for a geometric progression:$t_n = 2 + \frac{3(3^{n-1} - 1)}{3 - 1} = 2 + \frac{3^n - 3}{2} = \frac{4 + 3^n - 3}{2} = \frac{3^n + 1}{2}$Now,the sum of $n$ terms $S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} \frac{3^k + 1}{2} = \frac{1}{2} \left( \sum_{k=1}^{n} 3^k + \sum_{k=1}^{n} 1 \right)$$S_n = \frac{1}{2} \left( \frac{3(3^n - 1)}{3 - 1} + n \right)$$S_n = \frac{1}{2} \left( \frac{3(3^n - 1)}{2} + n \right) = \frac{n}{2} + \frac{3}{4}(3^n - 1)$

What is the sum of $n$ terms of the series $2 + 5 + 14 + 41 + \dots$?

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