1. ## Induction...gulp....

I need to prove the sume of a finite geometric series formula inductively...

summation (i=0 to n-1) ar^i = a((r^n)-1)/(r-1) for r not equal to 1 and n greater than or equal to one.

I know I need to show this for the base case and then for n implies n+1, but I'm stumped here.

2. Originally Posted by zhupolongjoe
I need to prove the sume of a finite geometric series formula inductively...

summation (i=0 to n-1) ar^i = a((r^n)-1)/(r-1) for r not equal to 1 and n greater than or equal to one.

I know I need to show this for the base case and then for n implies n+1, but I'm stumped here.
You need to prove that, for fixed $\displaystyle r\, \neq\, 1$, $\displaystyle r\, >\, 0$, and $\displaystyle n\, \geq\, 1$, the following holds:

. . . . .$\displaystyle \sum_{i=0}^{n-1} ar^i\, =\, \frac{a(r^n\, -\, 1)}{r\, -\, 1}$

I'm not sure why you're having trouble with the base case...? Just plug "1" in for "n":

. . . . .$\displaystyle \sum_{i=0}^{1-1} ar^i\, =\, \sum_{i=0}^{0} ar^0\, =\, a(1)\, -\, a$

...and:

. . . . .$\displaystyle \frac{a(r^1\, -\, 1)}{r\, -\, 1}\, =\, \frac{a(r\, -\, 1)}{r\, -\, 1}\, =\, a$

Where did you get stuck?

Then you assume:

. . . . .$\displaystyle \sum_{i=0}^{k-1} ar^i\, =\, \frac{a(r^k\, -\, 1)}{r\, -\, 1}$

Then you work with the formula with n = k + 1:

. . . . .$\displaystyle \sum_{i=0}^{(k+1)-1} ar^i\, =\, \sum_{i=0}^{(k-1)+1} ar^i\, =\, \sum_{i=0}^{k-1} ar^i\, +\, ar^k$

You can use the assumption step to plug in for the first part:

. . . . .$\displaystyle \frac{a(r^k\, -\, 1)}{r\, -\, 1}\, +\, ar^k$

. . . . .

3. Thanks, but your base case yields a=0.......that doesn't make sense to me.

4. Where is anything said about the value of $\displaystyle a$...?

5. You have the summation = a(1)-a, which is zero and then you have the other side as a(r-1)/(r-1)=a....I assume you meant a(1)=a and not -a.....

6. Sorry: I fat-fingered the keys, hitting the "-" instead of the "=" right next door. It should read:

Originally Posted by stapel
...the base case...? Just plug "1" in for "n":

. . . . .$\displaystyle \sum_{i=0}^{1-1} ar^i\, =\, \sum_{i=0}^{0} ar^0\, =\, a(1)\, =\, a$

...and:

. . . . .$\displaystyle \frac{a(r^1\, -\, 1)}{r\, -\, 1}\, =\, \frac{a(r\, -\, 1)}{r\, -\, 1}\, =\, a$

7. That's what I thought, okay thanks again.