Sixey Series

Calculus Level 2

Find the value of the infinite series

1 + 5 6 + 19 6 2 + 65 6 3 + 211 6 4 + … 1 + \frac{5}{6} + \frac{ 19}{6^2} + \frac{ 65} { 6^3} + \frac{ 211} { 6^4} + \ldots

Note: The pattern in the numerators is not immediately obvious. Bert believes that you can figure it out.


The answer is 3.

Bert Briones by Bert Briones , 50, Malaysia

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1 solution

The nth term of the series is of the form

( 3 n − 2 n ) / 6 n − 1 = 6 ∗ ( ( 1 / 2 ) n − ( 1 / 3 ) n ) (3^{n} - 2^{n}) / 6^{n-1} = 6 * ((1/2)^{n} - (1/3)^{n}) .

When the terms are then summed to infinity we end up with a result of

6 ∗ ( ∑ n = 1 ∞ ( 1 / 2 ) n − ∑ n = 1 ∞ ( 1 / 3 ) n ) = 6 ∗ ( 1 − ( 1 / 2 ) ) = 3 6 * (\sum_{n=1}^\infty(1/2)^{n} - \sum_{n=1}^\infty(1/3)^{n}) = 6 * (1 - (1/2)) = \boxed{3} .

(Note that I determined the formula for the nth term just by experimenting for a few minutes; indeed, the pattern wasn't obvious at first and I probably just got lucky finding it before too long.)

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