Find the sum of the series.

**deezy** · July 20th 2011, 10:09 PM

$\sum_{k = 0}^\infty \frac{3^{k-1}}{4^{3k+1}}$
$\sum_{k = 0}^\infty \frac{3^{k}*3^{-1}}{4^{3k}*4^{1}}$
$\sum_{k = 0}^\infty \frac{3^{k}}{(4^{3})^{k}*4^{1}*3^{1}}$
$(\frac{1}{12})\sum_{k = 0}^\infty \frac{3^{k}}{64^{k}}$
$(\frac{1}{12})\sum_{k = 0}^\infty (\frac{3}{64})^{k}$
So it's a geometric series with $a = 1$ and $r = 3/64$ . So $\frac{1}{12}*\frac{1}{1 - \frac{3}{64}}$ = $\frac{16}{183}$ ?

The solutions manual had $\frac{3}{15,616}$ . Am I doing this right?

**Prove It** · July 20th 2011, 10:18 PM

What you have done looks fine. Does your solutions manual post their full solution or just the answer?

**deezy** · July 20th 2011, 10:27 PM

There was a full solution, but I only copied down $\frac{3}{15,616}$ . I may have copied down the wrong answer... and I won't have a chance to check until tomorrow.

**deezy** · July 21st 2011, 01:27 PM

Well, the solutions manual had:
$\sum_{k = 2}^\infty \frac{3^{k-1}}{4^{3k+1}}$
$\sum_{k = 0}^\infty \frac{3^{k+1}}{4^{3k+7}}$
$(\frac{3}{4^7})\sum_{k = 0}^\infty (\frac{3}{4^{3}})^k$
$\frac{3}{4^7}*\frac{1}{1 - \frac{3}{4^3}}$
$\frac{3}{15616}$ .

But $\sum_{k = 2}^\infty \frac{3^{k-1}}{4^{3k+1}}$ was not the original problem, $\sum_{k = 0}^\infty \frac{3^{k-1}}{4^{3k+1}}$ was. So, pretty safe to say that's a typo?

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