Series in similar numbers-TIFR 2019 GS Part A, Problem 1

Understand the problem

The following sum of numbers (expressed in decimal notation)
\(1 + 11 + 111 +\cdots + \underbrace{11\cdots1}_{n \text{ terms}}\) is equal to (a)\((10^{n+1}-10-9n)/81 \) (b)\((10^{n+1}-10 + 9n)/81\) (c)\((10^{n+1}-10-n)/81\) (d)\((10^{n+1}-10 + n)/81\)

Source of the problem
TIFR GS 2019 Part A, Problem 1

Topic
Sequence and Progression

Difficulty Level
Easy

Suggested Book

Introduction to Real Analysis by Donald R. Sherbert and Robert G. Bartle

 

Start with hints

Do you really need a hint? Try it first!

\(S=\underbrace{(1+\cdots+1)}_{n \text{ terms}}+\underbrace{(1+\cdots+1)10}_{n-1 \text{ terms}}+\underbrace{(1+\cdots+1)10^2}_{n-2 \text{ terms}}+\cdots+1\times10^{n-1}\) where the sum in the first, second, third, \(\cdots\) brackets represent the sum of the digitss in the n, \(n-1, \cdots, 2,1 \) columns respectively in \( 11111111\cdots1 \\+1111111\cdots1 \\+\quad\quad111111\cdots1 \\+\quad\quad\quad11111\cdots1 \\+\quad\quad\quad\quad1111\cdots1 \\+\cdots\cdots\cdots \\+\cdots\cdots\cdots \\+\cdots\cdots\cdots \\+\quad\quad\quad\quad\quad\quad\quad\quad111 \\+\quad\quad\quad\quad\quad\quad\quad\quad\quad11 \\+\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad1 \) \(\Rightarrow S=n+(n-1)10+(n-2)10^2+\cdots+1\times10^{n-1}\)

\(10S=10n+(n-1)10^2+(n-2)10^3+\cdots+1\times10^{n}\)

\(S=n+(n-1)10+(n-2)10^2+\cdots+1\times10^{n-1}\) ;using these two equations what can you conclude about the sum \(S\)?

Substracting we get \(9S=-n+10+10^2+\cdots+1\times10^{n-1}+10^n\). This can be put in one of the form given as an options in the question.  

\( S=\frac{10^{n+1}-10-9n}{81}\) Hence, the solution.

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