The sum of the series 1 + frac{4}{3} + frac{1 | vedclass

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

Question

(C) The given series is $S_n = 1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots + n 	ext{ terms}$.We can rewrite the terms as:$S_n = (1) + (1 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given series is $S_n = 1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots + n 	ext{ terms}$.We can rewrite the terms as:$S_n = (1) + (1 + \frac{1}{3}) + (1 + \frac{1}{9}) + (1 + \frac{1}{27}) + \dots + n 	ext{ terms}$.Grouping the terms,we get:$S_n = (1 + 1 + 1 + \dots + n 	ext{ times}) + (\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots + n 	ext{ terms})$.$S_n = n + \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}}$.$S_n = n + \frac{\frac{1}{3}(1 - \frac{1}{3^n})}{\frac{2}{3}}$.$S_n = n + \frac{1}{2}(1 - \frac{1}{3^n})$.$S_n = n + \frac{1}{2} - \frac{1}{2 \cdot 3^n}$.

The sum of the series $1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \dots$ up to $n$ terms is

Similar Questions

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

Let $a, b, c, d$ be real numbers such that $\sum_{k=1}^n (a k^3+b k^2+c k+d)=n^4$,for every natural number $n$. Then,$|a|+|b|+|c|+|d|$ is equal to

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

Find the $9^{\text{th}}$ term in the sequence whose $n^{\text{th}}$ term is given by $a_{n} = (-1)^{n-1} n^{3}$.

$\sum\limits_{i=1}^n \sum\limits_{j=1}^i \sum\limits_{k=1}^j 1 = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module