The sum to infinity of the following series 2 | vedclass

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

Question

(C) The given series is $S = 2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots \infty$. We can rea

Vedclass · Accepted Answer

$7/2$

Vedclass · Answer

(C) The given series is $S = 2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots \infty$. We can rearrange the terms as: $S = (1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \infty) + (1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots \infty)$. Both parts are infinite geometric series with the sum formula $S_{\infty} = \frac{a}{1-r}$. For the first series,$a = 1$ and $r = 1/2$,so $S_1 = \frac{1}{1 - 1/2} = 2$. For the second series,$a = 1$ and $r = 1/3$,so $S_2 = \frac{1}{1 - 1/3} = \frac{1}{2/3} = 3/2$. Thus,the total sum is $S = S_1 + S_2 = 2 + 3/2 = 7/2$.

The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ is:

Similar Questions

In the sequence of sets $(1,2,3), (4,5,6), (7,8,9,10), \ldots$,the sum of elements in the $50^{th}$ set is

Let $\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots$ up to $10$ terms and $\beta = \sum_{n=1}^{10} n^4$. If $4\alpha - \beta = 55k + 40$,then $k$ is equal to . . . . . . .

If $\sum_{r=1}^{10} r! (r^3 + 6r^2 + 2r + 5) = \alpha(11!)$,then the value of $\alpha$ is equal to ...... .

Let $u_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$,for $n = 0, 1, 2, ...$. Then which of the following is true?

The sum to $n$ terms of the series $1^2 + (1^2 + 3^2) + (1^2 + 3^2 + 5^2) + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module