Find the sum of the series 1 + 3 + 7 + 15 + 3 | vedclass

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

Question

(B) Let the $n^{th}$ term be $T_n$ and the sum of $n$ terms be $S_n$.$S_n = 1 + 3 + 7 + 15 + \dots + T_n \quad (i)$$S_n = 1 + 3 + 7 + \dots + T_{n-1}

Vedclass · Accepted Answer

$2^{n+1} - n - 2$

Vedclass · Answer

(B) Let the $n^{th}$ term be $T_n$ and the sum of $n$ terms be $S_n$.$S_n = 1 + 3 + 7 + 15 + \dots + T_n \quad (i)$$S_n = 1 + 3 + 7 + \dots + T_{n-1} + T_n \quad (ii)$Subtracting $(ii)$ from $(i)$:$0 = 1 + [2 + 4 + 8 + 16 + \dots + (T_n - T_{n-1})] - T_n$$T_n = 1 + (2 + 4 + 8 + \dots + 2^{n-1})$$T_n = 1 + \frac{2(2^{n-1} - 1)}{2 - 1} = 1 + 2^n - 2 = 2^n - 1$Now,$S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (2^k - 1) = \sum_{k=1}^n 2^k - \sum_{k=1}^n 1$$S_n = \frac{2(2^n - 1)}{2 - 1} - n = 2^{n+1} - 2 - n$

Find the sum of the series $1 + 3 + 7 + 15 + 31 + \dots$ up to $n$ terms.

Similar Questions

If $\operatorname{gcd}(m, n) = 1$ and $1^2 - 2^2 + 3^2 - 4^2 + \ldots + (2021)^2 - (2022)^2 + (2023)^2 = 1012 m^2 n$,then $m^2 - n^2$ is equal to

The value of $\frac{1 \times 2^2 + 2 \times 3^2 + \ldots + 100 \times 101^2}{1^2 \times 2 + 2^2 \times 3 + \ldots + 100^2 \times 101}$ is

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$

The sum of the $1^{st} n$ terms of the series $\frac{1^{2}}{1} + \frac{1^{2}+2^{2}}{1+2} + \frac{1^{2}+2^{2}+3^{2}}{1+2+3} + \ldots$ is:

The sum of $1^3 + 2^3 + 3^3 + 4^3 + \dots + 15^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module