# 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

Easy

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

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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