From fe770f57ede4401b1c0ddba29fef658250b4c5fa Mon Sep 17 00:00:00 2001 From: arc Date: Wed, 1 Oct 2025 12:34:00 -0600 Subject: [PATCH] vault backup: 2025-10-01 12:34:00 --- education/math/MATH1220 (calc II)/Sequences.md | 7 +++++-- 1 file changed, 5 insertions(+), 2 deletions(-) diff --git a/education/math/MATH1220 (calc II)/Sequences.md b/education/math/MATH1220 (calc II)/Sequences.md index 99e6a08..a8a0075 100644 --- a/education/math/MATH1220 (calc II)/Sequences.md +++ b/education/math/MATH1220 (calc II)/Sequences.md @@ -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. 1. Rewrite $7^{-n}$ as $(\frac{1}{7})^n$ to be closer to the form $ar^n$ -$$ = \sum_{n=1}^\infty 35*($$ -2. \ No newline at end of file +$$ = \sum_{n=1}^\infty 35*(\frac{1}{7})^n*2^{n-1}$$ +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: +$$\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1}$$ \ No newline at end of file