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# 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!

$$S=\underbrace{(1+\cdots+1)}_{n \text{ terms}}+\underbrace{(1+\cdots+1)10}_{n-1 \text{ terms}}+\underbrace{(1+\cdots+1)10^2}_{n-2 \text{ terms}}+\cdots+1\times10^{n-1}$$ $$\Rightarrow S=n+(n-1)10+(n-2)10^2+\cdots+1\times10^{n-1}$$

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

Substracting we get $$9S=-n+10+10^2+\cdots+1\times10^{n-1}+10^n$$

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

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