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FP1 Series

How does equal this? -- when finfding the sum of the series

∑r=1n3r=32(3n−1)\displaystyle\sum_{r=1}^n 3^r = \frac{3}{2}(3^n-1)

I understand stand it when it just r or r sqaured, cubed etc.., but what do you do when it's to the power of r? Whats the standard result?
Reply 1
Avatar for around
around
13
Write out a few terms.

This should be a series you recognise...
Reply 2
Avatar for Narik
Narik
13
emmaxoxo
How does equal this? -- when finfding the sum of the series

∑r=1n3r=32(3n−1)\displaystyle\sum_{r=1}^n 3^r = \frac{3}{2}(3^n-1)

I understand stand it when it just r or r sqaured, cubed etc.., but what do you do when it's to the power of r? Whats the standard result?


Is this chapter 5 or chapter 6 - proof by induction? :s-smilie:
Reply 3
Avatar for oliv3r
oliv3r
1
emmaxoxo
How does equal this? -- when finfding the sum of the series

∑r=1n3r=32(3n−1)\displaystyle\sum_{r=1}^n 3^r = \frac{3}{2}(3^n-1)

I understand stand it when it just r or r sqaured, cubed etc.., but what do you do when it's to the power of r? Whats the standard result?


The standard result is:
∑r=1nar=nn−1(an−1)\displaystyle\sum_{r=1}^n a^r = \frac{n}{n-1}(a^n-1)
A similar result that might be useful is:
∑r=0nar=1n−1(an+1−1)\displaystyle\sum_{r=0}^n a^r = \frac{1}{n-1}(a^{n+1}-1)
I'm pretty sure you could prove these with induction if you needed to.

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