Find the sum of the series

[never_lose]

a = 1
r = 3/64
so,

=

The answer is supposed to be

Don't know why my answer is off. Am I doing this right? (Or maybe I copied down the wrong solution.)

 

[galactus]

Your answer is correct.

Maybe you copied the wrong solution?.

 

[tkhunny]

I'm going with "copied it down wrong". In any case, you seem to have the idea. Good work, except perhaps on the transcription.

 

[never_lose]

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

But \(\displaystyle \sum_{k = 2}^\infty \frac{3^{k-1}}{4^{3k+1}}\) was not the original problem, \(\displaystyle \sum_{k = 0}^\infty \frac{3^{k-1}}{4^{3k+1}}\) was. So, typo maybe? But thanks for the help, at least I know I'm doing these right.

 

[tkhunny]

Playing with the index should not change the value.

 

[renegade05]

Playing with the index should not change the value.

Would it not change the value for a though?

Since "a" is the first value of the series, the starting point would matter.

\(\displaystyle Sum: \frac{a}{1-r}\)

no ?? :?

 

[tkhunny]

Changing the starting value would change the sum. Playing with the index would not change the starting value without also adjusting the entire thing with an appropriate constant.

\(\displaystyle \frac{3^{k-1}}{4^{3k+1}} =\frac{3^{k}}{3\cdot 4\cdot 4^{3k}} = \frac{1}{12}\frac{3^{k}}{64^{k}}\)

k = 0

\(\displaystyle \frac{3^{0-1}}{4^{3(0)+1}} = \frac{1}{12}\)

\(\displaystyle \frac{1}{12}\cdot \frac{3^{0}}{64^{0}} = \frac{1}{12}\)