Series and sequencing...
- Thread starter hammy
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renegade05
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- Sep 10, 2010
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First is this your series?
\(\displaystyle \sum_{k=1}^{\infty}\frac{4^{k+1}}{9^{k-1}}\)
Second, what does the question ask for? What the sum of the series is?
If that is the case try expressing your series as the following:
\(\displaystyle \sum_{k=1}^{\infty}\frac{4^{k+1}}{9^{k-1}}\) \(\displaystyle \ = \ \sum_{k=1}^{\infty}\frac{4^k4^1}{\frac{9^k}{9^1}}\) \(\displaystyle = \ 36 * \sum_{k=1}^{\infty}(\frac{4}{9})^k\)
The last infinite series should look very familiar.
Hope this helps.
\(\displaystyle \sum_{k=1}^{\infty}\frac{4^{k+1}}{9^{k-1}}\)
Second, what does the question ask for? What the sum of the series is?
If that is the case try expressing your series as the following:
\(\displaystyle \sum_{k=1}^{\infty}\frac{4^{k+1}}{9^{k-1}}\) \(\displaystyle \ = \ \sum_{k=1}^{\infty}\frac{4^k4^1}{\frac{9^k}{9^1}}\) \(\displaystyle = \ 36 * \sum_{k=1}^{\infty}(\frac{4}{9})^k\)
The last infinite series should look very familiar.
Hope this helps.