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Math Help - Induction...gulp....

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    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 r\, \neq\, 1, r\, >\, 0, and n\, \geq\, 1, the following holds:

    . . . . . \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":

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

    ...and:

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

    Where did you get stuck?

    Then you assume:

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

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

    . . . . . \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:

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

    Then see where that leads.

    . . . . .
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  3. #3
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    Thanks, but your base case yields a=0.......that doesn't make sense to me.
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  4. #4
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    Where is anything said about the value of a...?
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  5. #5
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    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.....
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  6. #6
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    Sorry: I fat-fingered the keys, hitting the "-" instead of the "=" right next door. It should read:

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

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

    ...and:

    . . . . . \frac{a(r^1\, -\, 1)}{r\, -\, 1}\, =\, \frac{a(r\, -\, 1)}{r\, -\, 1}\, =\, a
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  7. #7
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    That's what I thought, okay thanks again.
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