The sum of the series 3 + 33 + 333 + dots + n | vedclass

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

Question

(B) Let the sum be $S_n = 3 + 33 + 333 + \dots + n$ terms.We can write $S_n = 3(1 + 11 + 111 + \dots + n 	ext{ terms})$.Multiply and divide by $9$:$S

Vedclass · Accepted Answer

$\frac{1}{27}(10^{n + 1} - 9n - 10)$

Vedclass · Answer

(B) Let the sum be $S_n = 3 + 33 + 333 + \dots + n$ terms.We can write $S_n = 3(1 + 11 + 111 + \dots + n 	ext{ terms})$.Multiply and divide by $9$:$S_n = \frac{3}{9}(9 + 99 + 999 + \dots + n 	ext{ terms})$$S_n = \frac{1}{3}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n 	ext{ terms})]$The first part is a geometric progression with $a = 10$,$r = 10$,and $n$ terms. The sum is $\frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$.$S_n = \frac{1}{3} \left[ \frac{10(10^n - 1)}{9} - n ight]$$S_n = \frac{1}{3} \left[ \frac{10^{n+1} - 10 - 9n}{9} ight]$$S_n = \frac{1}{27}(10^{n+1} - 9n - 10)$.

The sum of the series $3 + 33 + 333 + \dots + n$ terms is

Similar Questions

If $(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + ... + 10(11)^9 = k(10)^9$,then $k$ is equal to:

If $\tan n\theta = \tan m\theta$,then the different values of $\theta$ will be in

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

If $x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta$ and $y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta,$ for $0 < \theta < \frac{\pi}{4},$ then

If $\alpha, \beta$ are the roots of the equation $x^2 - 3x + a = 0$ and $\gamma, \delta$ are the roots of the equation $x^2 - 12x + b = 0$,and $\alpha, \beta, \gamma, \delta$ form an increasing $G.P.$,then $(a, b) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module