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The sum of $n$ terms of the series $4+44+444+ \dots \dots $ is

  1. $\frac{4}{81}\left[10^{n+1}-9n-1\right]$
  2. $\frac{4}{81}\left[10^{n-1}-9n-1\right]$
  3. $\frac{4}{81}\left[10^{n+1}-9n-10\right]$
  4. $\frac{4}{81}\left[10^{n}-9n-10\right]$
in Numerical Ability by Veteran (59.4k points)
edited by | 326 views
+2

take only 4 and put n=1

option C sum =4   :-P

3 Answers

+5 votes
Best answer

The sum of n terms of the series $4+44+444+ \ldots$ is

Final sum $= 4(1+11+111+1111+\ldots)$

Let $S_n = \underbrace{1+11+111+1111+\ldots +1111}_{n \text{ times}}$

∴ $S_0 = 0$ and $S_1 = 1$ and $S_2 = 1+11 = 12.$

$S_n = (10^0)+ (10^1+1)+(10^2+11)+(10^3+111)+\ldots+ \left(10^{n-1}+\underbrace{111\ldots 1}_{n-1 \text{ times}}\right)$

$\implies S_n = ({10^0+10^1+10^2+10^3+\ldots10^{n-1}} )+ \left({1+11+111+1111+\ldots+\underbrace{111\ldots 1}_{n-1 \text{ times}} }\right)$

  • $\implies S_n = \left (\frac{10^{n}-1}{9} \right)+ (S_{n-1})\quad \to (1)$

Recursively substituting, we get

$S_n = \left(\frac{10^{n}-1}{9} \right)+ \left(\frac{10^{n-1}-1}{9} \right)+\left(\frac{10^{n-2}-1}{9} \right)+\ldots+\left(\frac{10^{1}-1}{9} \right)+(S_{0})$ 

$\implies S_n = \frac{1}{9}\left[\left(10^{n}-1 \right)+ \left(10^{n-1}-1\right )+\left(10^{n-2}-1\right )+\ldots+\left(10^{1}-1\right ) \right]$

$\implies S_n = \frac{1}{9}\left[10^{n}+ 10^{n-1}+10^{n-2}+\ldots+10^{1}- n.1 \right]$

$\implies S_n = \frac{1}{9}\left[10^{1}+ 10^{2}+10^{3}+\ldots+10^{n}- n \right]$

$\implies S_n = \frac{1}{9}\left[10\left(\frac{10^{n}-1}{9}\right)- n \right] = \frac{1}{9\times 9}\left[10^{n+1}-10-9 n \right] = \frac{1}{81}\left[10^{n+1}-9n-10 \right]$

Final SUM $= 4 S_n = \frac{4}{81}\left[10^{n+1}-10-9 n \right] = \frac{4}{81}\left[10^{n+1}-9n-10 \right]$

Shortcut :-

in the question, they asked for sum of $n$ terms.

So, let $n=3$

$\implies$ Required sum $= 4+44+444 = 492$

Put in each option $n=3,$ it matches only for Option C.

(If more than one option is matched, then take $n=4,n=5$ until you get only one option is matched!)

Correct Answer: Option C.

by Veteran (65.8k points)
edited by
+3
There is a bit mistake. It may be a typing one. $10$ is missing from the expression of the $5^{\mathrm{th}}$ line of $S_n$

In that line it will be
$ S_n=\frac{1}{9}[10\left( \frac{10^n-1}{9} \right)-n]$
+3 votes
Let us suppose $S = 4+44+444+ \dots \dots\: n\:\text{times}$

$\implies S = 4\left [1+11+111+ \dots \dots \: n\:\text{times} \right]$

$\implies S = \dfrac{4}{9}\left [9+99+999+ \dots \dots \: n\:\text{times} \right]$

$\implies S = \dfrac{4}{9}\left [(10-1)+(10^{2}-1)+(10^{3}-1)+ \dots \dots \: n\:\text{times} \right]$

$\implies S = \dfrac{4}{9}\left [\underbrace{10^{1} + 10^{2} + 10^{3}+ \dots \dots \: n\:\text{times}}_{\textbf{GP Series}}\:- n \right]$

$\implies S = \dfrac{4}{9}\left[\dfrac{10(10^{n} - 1)}{10-1} - n \right]\:\:\:\:\left[\because\text{Sum of GP}: S_{n} = \dfrac{a(r^{n} - 1)}{r-1}\:\:; r>0\:\: \text{and}\:\: S_{n} = \dfrac{a(1- r^{n})}{1-r}\:\:; r< 0 \right]$

$\implies S = \dfrac{4}{9}\left[\dfrac{10^{n+1} - 10-9n}{9}\right]$

$\implies S = \dfrac{4}{81}\left[10^{n+1} - 9n - 10 \right]$

So, the correct answer is $(C).$
by Veteran (59.4k points)
edited by
0 votes
Sum of 1 term=4; Sum of Two term=4+44; ......

Now if we put n=1 in options we get 4 as answer

A.4/81*90     B.-4/81*9     C.4     D.-4/81*9

So. C is correct
by (175 points)
Answer:

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