Finite geometric series formula (video) | Khan Academy

Want to join the conversation?Log inSort by:averynash8 years agoPosted 8 years ago. Direct link to averynash's post “At 4:25, Sal multiplies a...”At 4:25, Sal multiplies ar^(n-1) by -r and gets -ar^n. I do not quite get how that works and would like some help on it. Thanks in advance!Button navigates to signup page•Button navigates to signup page(67 votes)yeloc18 years agoPosted 8 years ago. Direct link to yeloc1's post “OK, this is a really REAL...”OK, this is a really REALLY great question. When you multiply ar^(n-1) and -r together the first thing you can do is distribute the negative sign, which gives you -ar^(n-1) * r. The variable r can also be expressed as r^1. So you get -ar^(n-1) * r^1. Next you can pull out the -a which gives you (-a)(r^(n-1)) * r^1. Then you can simplify and get (-a)(r^(n-1+1)). Once again that can be simplified very easily to -ar^n. I hope that was helpful.Comment on yeloc1's post “OK, this is a really REAL...”(135 votes)Eliza4 years agoPosted 4 years ago....