Sum of the series frac{2}{3} + frac{8}{9} + f | vedclass

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

Question

(D) The $n^{th}$ term of the series is given by $T_n = \frac{3^n - 1}{3^n} = 1 - \left(\frac{1}{3}\right)^n$. The sum of $n$ terms is $S_n = \sum_{k=1

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The $n^{th}$ term of the series is given by $T_n = \frac{3^n - 1}{3^n} = 1 - \left(\frac{1}{3}\right)^n$. The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n \left(1 - \left(\frac{1}{3}\right)^k\right)$. $S_n = \sum_{k=1}^n 1 - \sum_{k=1}^n \left(\frac{1}{3}\right)^k$. $S_n = n - \left[ \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} \right]$. $S_n = n - \frac{\frac{1}{3}(1 - 3^{-n})}{\frac{2}{3}} = n - \frac{1}{2}(1 - 3^{-n})$. $S_n = n - \frac{1}{2} + \frac{1}{2}3^{-n} = n + \frac{1}{2}(3^{-n} - 1)$.

Sum of the series $\frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{80}{81} + \dots$ to $n$ terms is

Similar Questions

Let $A = \sum_{i=1}^{10} \sum_{j=1}^{10} \min \{i, j\}$ and $B = \sum_{i=1}^{10} \sum_{j=1}^{10} \max \{i, j\}$. Then $A + B$ is equal to

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. Its sum is $4$ and the second term is $3/4$. Then:

The product $(32)(32)^{1/6}(32)^{1/36} \dots \infty$ is

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

$\sum\limits_{r = 1}^n {\sum\limits_{m = 1}^r {m} } = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module