Infinity infinity

Calculus Level 4

$\large S=\sum_{n=1}^{\infty} \dfrac{q_{n}}{3^{n}}$

Let the $q_{n}$ be a defined sequence for $q_{1}=1$ , $q_{2}=2$ , $q_{3}=3$ such that it satisfy the recurrence relation $q_{n+3}=q_{n+2}-q_{n+1}+q_{n}$ for positive integer $n$ .

Compute $S$ .

The answer is 0.7.

1 solution

Guilherme Niedu
May 4, 2016

The series is an alternating series in the form:

$\{1,2,3,2,1,2,3,2,1,2,3,2,1... \}$

Therefore:

$\large \displaystyle S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{2}{3^4} + \frac{1}{3^5} + ...$

Multiplying by 3:

$\large \displaystyle 3S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} + ...$

Subtracting:

$\large \displaystyle 2S = 1 + \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} - \frac{1}{3^4} + \frac{1}{3^5} + \frac{1}{3^6} + ...$

$\large \displaystyle 2S = 1 + (\frac{1}{3^2} - \frac{1}{3^4} + \frac{1}{3^6} - ... ) + (\frac{1}{3} - \frac{1}{3^3} + \frac{1}{3^5} -...)$

$\large \displaystyle 2S = 1 + \frac{\frac{1}{9}}{1 - \frac{-1}{9}} + \frac{\frac{1}{3}}{1 - \frac{-1}{9}}$

$\large \displaystyle 2S = 1 + \frac{1}{10} + \frac{3}{10} = \frac{14}{10}$

$\large \displaystyle S = \frac{7}{10} = \fbox{0.7}$

Infinity infinity

The answer is 0.7.

1 solution

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