Triple progress

Algebra Level 3

( 1 + 2 3 + ( 2 3 ) 2 + ⋯ ) + ( 1 + 3 4 + ( 3 4 ) 2 + ⋯ ) + ( 1 + 4 5 + ( 4 5 ) 2 + ⋯ ) = ? \left(1+ \frac 23 + \left(\frac 23\right)^2 +\cdots \right) + \left(1 + \frac 34 + \left(\frac 34 \right)^2 + \cdots \right) + \left(1 + \frac 45 + \left(\frac 45 \right)^2 + \cdots \right) = \, ?


The answer is 12.

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Denton Young by Denton Young , 47, USA

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2 solutions

Chew-Seong Cheong
Aug 27, 2016

Relevant wiki: Geometric Progression Sum

S = ( 1 + 2 3 + ( 2 3 ) 2 + . . . ) + ( 1 + 3 4 + ( 3 4 ) 2 + . . . ) + ( 1 + 4 5 + ( 4 5 ) 2 + . . . ) = ∑ n = 0 ∞ ( 2 3 ) n + ∑ n = 0 ∞ ( 3 4 ) n + ∑ n = 0 ∞ ( 4 5 ) n = 1 1 − 2 3 + 1 1 − 3 4 + 1 1 − 4 5 = 3 + 4 + 5 = 12 \begin{aligned} S & = \left(1+ \frac 23 + \left(\frac 23\right)^2 + ...\right) + \left(1 + \frac 34 + \left(\frac 34 \right)^2 + ...\right) + \left(1 + \frac 45 + \left(\frac 45 \right)^2 + ...\right) \\ & = \sum_{n=0}^\infty \left(\frac 23\right)^n + \sum_{n=0}^\infty \left(\frac 34\right)^n + \sum_{n=0}^\infty \left(\frac 45\right)^n \\ & = \frac 1{1-\frac 23} + \frac 1{1-\frac 34} + \frac 1{1-\frac 45} \\ & = 3 + 4 + 5 = \boxed{12} \end{aligned}

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Denton Young
Aug 26, 2016

By GP rule, first progression equals 3, second = 4, third = 5. 3 + 4 + 5 = 12

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