Anonymous at Sat, 27 Apr 2024 19:57:06 UTC No. 16148881

>>16148874
Multiply the series with [math]\frac{1-r}{1-r}[/math] and you get a telescoping series.

Anonymous at Sat, 27 Apr 2024 20:12:06 UTC No. 16148913

>>16148881
I ended up with a((1-r^n-r^(n+1))/(1-r)), which means I probably fucked up somewhere in the multiplication, thanks though

Anonymous at Sat, 27 Apr 2024 20:32:09 UTC No. 16148937

>>16148881
I don't get it, i get a r^(n+1) instead of r^n
I looked up telescoping series but this doesnt seem to have the same effect because if you are gonna multiply by something you will have to multiply everything in the parenthesis

Anonymous at Sat, 27 Apr 2024 21:37:35 UTC No. 16149011

>>16148874
Just do partial fractions

Anonymous at Sat, 27 Apr 2024 21:40:20 UTC No. 16149015

[eqn] \sum_{k=1}^n a r^{k-1} \\
= \frac{1 - r}{1 - r} \sum_{k=1}^n a r^{k-1} \\
= \frac{1}{1 - r} \sum_{k=1}^n a (r^{k - 1} - r^k) \\
= \frac{1}{1 - r} (a - a r^n) \\
= a \frac{1 - r^n}{1 - r}
[/eqn]

Anonymous at Sat, 27 Apr 2024 22:05:30 UTC No. 16149043

>>16149015 nice i get it now, thx

Anonymous at Sat, 27 Apr 2024 22:35:09 UTC No. 16149089

>>16148874
Take Sn and multiply both it by r. Then Sn-r*Sn=a--ar^n, Solving that for Sn gives you the result.