Find the sum of the following series up to n | vedclass

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

Question

(B) Let $S_n = 5 + 55 + 555 + \ldots$ to $n$ terms.$S_n = 5(1 + 11 + 111 + \ldots 	ext{ to } n 	ext{ terms})$$S_n = \frac{5}{9}(9 + 99 + 999 + \ldot

Vedclass · Accepted Answer

$\frac{50}{81}(10^n-1)-\frac{5n}{9}$

Vedclass · Answer

(B) Let $S_n = 5 + 55 + 555 + \ldots$ to $n$ terms.$S_n = 5(1 + 11 + 111 + \ldots 	ext{ to } n 	ext{ terms})$$S_n = \frac{5}{9}(9 + 99 + 999 + \ldots 	ext{ to } n 	ext{ terms})$$S_n = \frac{5}{9}((10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n-1))$$S_n = \frac{5}{9}((10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + 1 + \ldots + 1))$Using the sum formula for a geometric progression $a(r^n-1)/(r-1)$ where $a=10$ and $r=10$:$S_n = \frac{5}{9}\left(\frac{10(10^n-1)}{10-1} - night)$$S_n = \frac{5}{9}\left(\frac{10(10^n-1)}{9} - night)$$S_n = \frac{50}{81}(10^n-1) - \frac{5n}{9}$

Find the sum of the following series up to $n$ terms:
$5+55+555+\ldots$

Similar Questions

Let $a_{1}=b_{1}=1$,$a_{n}=a_{n-1}+2$,and $b_{n}=a_{n}+b_{n-1}$ for every natural number $n \geq 2$. Then $\sum_{n=1}^{15} a_{n} \cdot b_{n}$ is equal to $.........$

Let $S_{k} = \frac{1+2+\ldots+k}{k}$ and $\sum_{j=1}^n S_j^2 = \frac{n}{A}(Bn^2 + Cn + D)$,where $A, B, C, D \in \mathbb{N}$ and $A$ has the least value. Then:

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

The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots$ is equal to

If $\sum\limits_{r=1}^\infty \frac{1}{(2r-1)^2} = \frac{\pi^2}{8}$,then $\sum\limits_{r=1}^\infty \frac{1}{r^2} = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the sum of the following series up to $n$ terms:$5+55+555+\ldots$