Find the sum of n terms of the series 1 + 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 $t_n$. The series is $1, 3, 7, 15, 31, \dots$The terms can be written as $(2^1 - 1), (2^2 - 1), (2^3 - 1), (2^4 - 1), (2^5

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the $n$-th term be $t_n$. The series is $1, 3, 7, 15, 31, \dots$The terms can be written as $(2^1 - 1), (2^2 - 1), (2^3 - 1), (2^4 - 1), (2^5 - 1), \dots$Thus,the $n$-th term is $t_n = 2^n - 1$.The sum of $n$ terms $S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} (2^k - 1)$.$S_n = \sum_{k=1}^{n} 2^k - \sum_{k=1}^{n} 1$.The first part is a geometric progression with first term $a = 2$ and common ratio $r = 2$.Sum of $n$ terms of $GP$ $= \frac{a(r^n - 1)}{r - 1} = \frac{2(2^n - 1)}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2$.Therefore,$S_n = (2^{n+1} - 2) - n = 2^{n+1} - n - 2$.

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

Similar Questions

The sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n + 1)^2}{2}$ when $n$ is even. When $n$ is odd,the sum is:

What is the sum of the infinite series $1^2 + 2^2 x + 3^2 x^2 + \dots$?

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 sum of the series $3.6 + 4.7 + 5.8 + \dots$ up to $(n - 2)$ terms is:

The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module