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education/math/MATH1220 (calc II)/Sequences.md
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@@ -78,5 +78,8 @@ Converges to $\dfrac{a}{1-r}$ if $|r| < 1$ or diverges if $|r| >= 1$.
> Determine if the series $\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1})$ diverges or converges. If it converges, state where. > Determine if the series $\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1})$ diverges or converges. If it converges, state where.
1. Rewrite $7^{-n}$ as $(\frac{1}{7})^n$ to be closer to the form $ar^n$ 1. Rewrite $7^{-n}$ as $(\frac{1}{7})^n$ to be closer to the form $ar^n$
$$ = \sum_{n=1}^\infty 35*($$ $$ = \sum_{n=1}^\infty 35*(\frac{1}{7})^n*2^{n-1}$$
2. 2. Pull a $7$ out of the bottom, making use of the fact that $(\frac{1}{7})^n = \frac{1}{7}(\frac{1}{7})^{n-1}$
$$ = \sum_{n=1}^\infty \frac{35}{7}(\frac{1}{7})^{n-1}* 2^{n-1} = \sum_{n = 1}^\infty \frac{35}{7}(\frac{2}{7})^{n-1} $$
3. This is now of the form $\sum_{n=1}^\infty ar^{n-1}$, so:
$$\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1}$$