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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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 $2 + 5 + 14 + 41 + \dots$?

Similar Questions

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

If $2^3+4^3+6^3+\ldots+(2n)^3 = h n^2(n+1)^2$,then $h$ is equal to

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

The odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:

If the $n^{th}$ term of a sequence is $n(n + 1)$,then what is the sum of its $n$ terms?

Vedclass Test Series

Exam Paper Generator

Online Exam Module