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Algebra Level 2

∑ n = 1 ∞ ( 2 1 0 2 n − 1 + 3 1 0 2 n ) = ? \sum_{n=1}^\infty \left(\frac 2{10^{2n-1}} + \frac 3{10^{2n}} \right) = \ ?

23 99 \frac{23}{99} 1 1 Undefined 17 99 \frac{17}{99} 32 99 \frac{32}{99}

Ethan Mandelez by Ethan Mandelez , 15, Malaysia

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2 solutions

Since the series converges, we can calculate the sum as follows:

S = ∑ n = 1 ∞ ( 2 1 0 2 n − 1 + 3 1 0 2 n ) = 2 10 + 3 1 0 2 + 2 1 0 3 + 3 1 0 4 + 2 1 0 5 + 3 1 0 6 + ⋯ 10 S = 2 + 3 10 + 2 1 0 2 + 3 1 0 3 + 2 1 0 4 + 3 1 0 5 + ⋯ 9 S = 2 + 1 10 − 1 1 0 2 + 1 1 0 3 − 1 1 0 4 + 1 1 0 5 − ⋯ 90 S = 20 + 1 − 1 10 + 1 1 0 2 − 1 1 0 3 + 1 1 0 4 − ⋯ 99 S = 23 ⟹ S = 23 99 \begin{aligned} S & = \sum_{n=1}^\infty \left(\frac 2{10^{2n-1}} + \frac 3{10^{2n}} \right) \\ & = \frac 2{10} + \frac 3{10^2} + \frac 2{10^3} + \frac 3{10^4} + \frac 2{10^5} + \frac 3{10^6} + \cdots \\ 10S & = 2 + \frac 3{10} + \frac 2{10^2} + \frac 3{10^3} + \frac 2{10^4} + \frac 3{10^5} + \cdots \\ 9S & = 2 + \frac 1{10} - \frac 1{10^2} + \frac 1{10^3} - \frac 1{10^4} + \frac 1{10^5} - \cdots \\ 90S & = 20 + 1 - \frac 1{10} + \frac 1{10^2} - \frac 1{10^3} + \frac 1{10^4} - \cdots \\ 99S & = 23 \\ \implies S & = \boxed{\frac {23}{99}} \end{aligned}

1 Helpful 0 Interesting 2 Brilliant 0 Confused
Ethan Mandelez
Dec 1, 2020

Here is one way of doing it:

3 Helpful 1 Interesting 2 Brilliant 0 Confused

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