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Math Help - 'Proof' help needed please...

  1. #1
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    'Proof' help needed please...

    I need to prove by induction on n, that

    4^1 + 4^2 + 4^3 +....+ 4^n = 4/3 (4^n) - 1

    For all integers 'n' is = or > than 1


    Thank you very much!
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  2. #2
    MHF Contributor chisigma's Avatar
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    The sum of the first n terms of a 'geometric progression' is...

    S_{n} = a_{0}\cdot (1 + r + r^{2} + ... r^{n-1}) = a_{0}\cdot \frac{r^{n}-1}{r-1} (1)

    Setting in (1) a_{0}= r = 4 what do You obtain?...

    Kind regards

    \chi \sigma
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  3. #3
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    Sorry, but I am totally lost!

    Oz
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  4. #4
    MHF Contributor chisigma's Avatar
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    Taking into account the identity...

    r^{n}-1 = (r-1)\cdot (1 + r + r^{2} + \dots + r^{n-1}) (1)

    ... you arrive to demostrate that is...

    S_{n} = a_{0}\cdot (1 + r + r^{2} + \dots + r^{n-1})= a_{0}\cdot \frac {r^{n}-1}{r-1} (2)

    ... so that...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by OsbourneOz View Post
    I need to prove by induction on n, that

    4^1 + 4^2 + 4^3 +....+ 4^n = 4/3 (4^n) - 1

    For all integers 'n' is = or > than 1
    In a proof by induction, you have to
    a) check the initial case: here, you have to check the identity when n=1.
    b) perform the induction step: assume that, for some n\geq 1, 4^1+\cdots+4^n=\frac{4}{3}4^n-1; then you have to prove that 4^1+4^2+\cdots+4^{n+1}=\frac{4}{3}4^{n+1}-1 (same identity with n+1 instead of n). To do this, simply note that 4^1+\cdots+4^{n+1}=(4^1+\cdots+4^n)+4^{n+1}=(\frac  {4}{3}4^n-1)+4^{n+1} (using the assumption), and after simplification you should get what you need. This will conclude the induction.
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