Mistakes give rise to problems.

Calculus Level 4

1 3 + 2 9 + 3 27 + 4 81 + 5 729 + 8 2187 + 7 6561 + … \frac { 1 }{ 3 } +\frac { 2 }{ 9 } +\frac { 3 }{ 27 } +\frac { 4 }{ 81 } +\frac { 5 }{ 729 } + \frac { 8 } { 2187} + \frac{ 7}{ 6561} + \ldots

i.e. Evaluate ∑ n = 1 ∞ ( ( 1 3 ) n + ( 2 9 ) n ) = a b \sum_{n=1}^{\infty} \left( \left(\dfrac{1}{3}\right)^n + \left(\dfrac{2}{9}\right)^n \right) = \dfrac {a}{b}

If a , b a,b are co-prime integers, calculate a b ab .

Inspiration


The answer is 154.

Mehul Arora by Mehul Arora , 20, India

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1 solution

Mehul Arora
Feb 17, 2016

We observe that the series is two interwoven GP's.

1 3 + 2 9 + 3 27 + 4 81 + 5 135 + 8 729 ⋯ = a b \dfrac {1}{3}+ \dfrac {2}{9}+\dfrac {3}{27}+ \dfrac {4}{81}+ \dfrac {5}{135}+ \dfrac {8}{729} \cdots= \dfrac {a}{b} =

1 3 + 2 9 + 1 9 + 4 81 + 1 27 + 8 729 ⋯ = a b \color{#D61F06}{\dfrac {1}{3}}+ \color{#007fff}{\dfrac {2}{9}}+ \color{#D61F06}{\dfrac {1}{9}}+ \color{#007fff}{\dfrac {4}{81}}+ \color{#D61F06}{\dfrac {1}{27}}+ \color{#007fff}{\dfrac {8}{729}} \cdots= \dfrac {a}{b}

= 1 3 1 − 1 3 + 2 9 1 − 2 9 = 1 2 + 2 7 = 11 14 = \dfrac {\dfrac{1}{3}}{1- \dfrac {1}{3}} + \dfrac {\dfrac {2}{9}}{1- \dfrac {2}{9}}= \dfrac {1}{2}+ \dfrac {2}{7}= \dfrac {11}{14}

a b = 154 \huge {ab=154}

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