The sum of 100 terms of the series 0.9+0.09+0.009+.... will be?

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Answer 1 · 2018-04-28T14:38:36

Sum $approx 1$

Sum $= 0.9+0.09+0.009+ ...$

The sum is a geometric progression (GP) with first term $(a_1)=0.9$ and common ratio $(r)=0.1$

The sum of the first $n$ terms of a GP is given by:

Sum $= (a_1(1-r^n))/(1-r)$

In this case,

Sum $= (0.9(1-0.1^100))/(1-0.1)$

$= (cancel0.9(1-0.1^100))/cancel0.9$

$approx (1-0)$

$approx 1$

[NB: $0.1^100 = 1 xx 10^-100$ so $(1- 0.1^100)$ is actually $0.$ followed by 100 $9$ 's]