## math_proof 3 years ago prove the same result, not by induction, but by directly manipulating the sum: let A be the sum, and show that xA = A + xn+1 -1.(Use sigma notation in your proof).

1. math_proof

$\sum_{i=0}^{n} x ^{i} = \frac{ 1-x ^{n+1} }{ 1-x }$

2. KingGeorge

First off, write (in sigma notation) what xA would be. What do you get?

3. math_proof

i don't get it, whats xA

4. KingGeorge

$A=\sum_{i=0}^n x^i$xA=?

5. math_proof

im so confused

6. math_proof

i hate does things

7. math_proof

those

8. KingGeorge

Maybe it would help if you wrote the sum out. So you would get $\sum_{i=0}^n x^i=x^0+x^1+x^2+x^3+...$Then, $xA=x\sum_{i=0}^n x^i=x(x^0+x^1+x^2+x^3+...)=x^1+x^2+x^3+x^4+...$Can you write this back in sigma notation?

9. math_proof

so thats gonna be the same as the above one

10. KingGeorge

Almost. There's one little difference. Your new sum should be $\sum_{i=0}^n x^{i+1}$All that's changed, is that each power of x has 1 added to it. Do you understand why this happens?

11. math_proof

yeah i get it, but can't you change the index to start from 1? wouldn't it be the same?

12. KingGeorge

You could, you just need to be careful to also set the final index to n+1.That's why I prefer to start at 0. So $\sum_{i=0}^n x^{i+1}=\sum_{i=1}^{n+1} x^i$Now, all you need to do, is show that $$A+x^{n+1}-1$$ is the same thing as that new sum.

13. math_proof

im having trouble understanding what the questions asks as to prove. whats A is the sum, then $x \sum_{0}x ^{i}$

14. math_proof

so its basically xA=A+xn+1, but what about -1?

15. KingGeorge

Notice that your new sum does not have the $$x^0=1$$ term. So you need to subtract off the 1.

16. math_proof

I don't get how its gonna equal $\frac{ 1-x ^{n+1} }{ 1-x }$, how does it prove that? if I basically get $\sum_{i=0}^{n} x ^{n+1}-1$

17. KingGeorge

Sorry it took so long for me to get back to you. OS was acting up for me quite seriously. But you want to show that $\sum_{i=0}^n x^{i+1}=\left(\sum_{i=0}^n x^i\right)+x^{n+1}-1$There shouldn't be any division involved.