The sum of the first 20 terms of the sequence | vedclass

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

Question

(C) Let $S_{20} = 0.7 + 0.77 + 0.777 + \dots$ up to $20$ terms. $S_{20} = 7[0.1 + 0.11 + 0.111 + \dots 	ext{ up to } 20 	ext{ terms}]$. Multiply and

Vedclass · Accepted Answer

$\frac{7}{81}(179 + 10^{-20})$

Vedclass · Answer

(C) Let $S_{20} = 0.7 + 0.77 + 0.777 + \dots$ up to $20$ terms. $S_{20} = 7[0.1 + 0.11 + 0.111 + \dots 	ext{ up to } 20 	ext{ terms}]$. Multiply and divide by $9$: $S_{20} = \frac{7}{9}[0.9 + 0.99 + 0.999 + \dots 	ext{ up to } 20 	ext{ terms}]$. $S_{20} = \frac{7}{9}[(1 - 10^{-1}) + (1 - 10^{-2}) + (1 - 10^{-3}) + \dots + (1 - 10^{-20})]$. $S_{20} = \frac{7}{9}[20 - (10^{-1} + 10^{-2} + \dots + 10^{-20})]$. The sum of the geometric series inside the bracket is $\frac{0.1(1 - 0.1^{20})}{1 - 0.1} = \frac{0.1(1 - 10^{-20})}{0.9} = \frac{1}{9}(1 - 10^{-20})$. $S_{20} = \frac{7}{9}[20 - \frac{1}{9}(1 - 10^{-20})] = \frac{7}{9}[\frac{180 - 1 + 10^{-20}}{9}] = \frac{7}{81}(179 + 10^{-20})$.

The sum of the first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$ is

Similar Questions

The sum of the first $n$ terms of the series $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ is

If $\alpha_r$ and $\beta_r$ (where $\alpha_r < \beta_r$) are the roots of the quadratic equation $x^2 - r^2(r + 1)x + r^5 = 0$,then find the value of $\sum_{r=1}^{n} (3\alpha_r + 2\beta_r)$.

For a series whose $n^{th}$ term is $\left( \frac{n}{x} \right) + y$,the sum of $r$ terms is:

The number of positive integers $n$ in the set $\{1, 2, 3, \ldots, 100\}$ for which the number $\frac{1^2+2^2+3^2+\ldots+n^2}{1+2+3+\ldots+n}$ is an integer is

The sum of the series $6 + 66 + 666 + \dots$ up to $n$ terms is

Vedclass Test Series

Exam Paper Generator

Online Exam Module