TIFR CSE 2011 | Part A | Question: 8

Question

The sum of the first $n$ terms of the series $1, 11, 111, 1111,\dots,$ is.

• A. $\frac{1}{81}\left ( 10^{n+1}-9n-10 \right )$
• B. $\frac{1}{81}\left ( 10^{n}-9n \right )$
• C. $\frac{1}{9}\left ( 10^{n+1}-1\right )$
• D. $\frac{1}{9}\left ( 10^{n+1}-n10^{n}\right )$
• E. None of the above

Answer 1

$S = (1+11+111 + \cdots n \text{ terms})$

$=\dfrac 19 \times (9+99+999 + \cdots n \text{ terms)}$

$=\dfrac19 \Bigl ((\color{blue}{10}-\color{red}{1})+(\color{blue}{100}-\color{red}{1})+(\color{blue}{1000}-\color{red}{1}) + \cdots n \text{ terms} \Bigr )$

$= \dfrac19 \Bigl (\color{blue}{(10+100 + \cdots 10^n)} - \color{red}{(1+1+\cdots n \text{ terms)}} \Bigr )$

$=\dfrac 19 \left ( \color{blue}{\dfrac{10^{n+1} - 10}{10-1}} - \color{red}{n} \right )$

$=\dfrac 19 \left ( \dfrac{10^{n+1} - 10 - 9n}{9} \right )$

$=\dfrac{10^{n+1} - 9n - 10}{81}$

So, the correct answer is option A.

Comments

This type of question easily solves by elimination method.

i.e. if n = 1 ,then sum = 1

put the  n =1 in the given options and easily get the answer.

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bro that's a different thing to reach the answer by option and get the answer  by approach many have done this just by putting value and eliminate options she would have done the same and that's a good thing in exam where you have less time, but here we are learning & building our concept so "approach matters" you should have "appreciate" her effort for adding nice explanation.

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Yes

the approach is better to know.

Very Nice Explanation

@Pooja Palod ma'am Thank you so much