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Math Help - Mathematical induction

  1. #1
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    Mathematical induction

    Please could anyone help!

    I need to use mathematical induction to prove that

    1+5+5^2+...+5^n=1/4(5^(n+1)-1)

    for all positive intergers n


    Thanks
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  2. #2
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    For \displaystyle  1+5+5^2+\dots +5^n=\frac{1}{4}(5^{n+1}-1)

    First step

    Make \displaystyle  f(n) = 1+5+5^2+\dots +5^n

    Show \displaystyle  f(1)
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  3. #3
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    Thanks i did the first step, its the second part I am stuck on, the K+1 bit.

    If you could offer any help I would be very appreciative.

    Thanks.
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Make \displaystyle  f(n) = 1+5+5^2+\dots +5^n

    Show \displaystyle  f(1)
    To be precise, one must define a property, not a function. If f:\mathbb{N}\to\mathbb{N}, then f(1) is neither true nor false, so one cannot prove f(1). Instead, define P(n) to be \displaystyle  1+5+5^2+\dots +5^n=\frac{1}{4}(5^{n+1}-1) and prove P(1) and \forall k\,(P(k)\to P(k+1)).

    Quote Originally Posted by Arron
    its the second part I am stuck on, the K+1 bit.
    Why don't you post the induction hypothesis, what you need to prove and what exactly your problem is?
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  5. #5
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    This is how far I have got,

    P(1) is true, since

    1+5^1=6 and 1/4(5^1+1 -1) =6

    Now let k>1, and assume that P(k) is true, that is

    1+5+5^2+...+5^K = 1/4(5^k+1 -).

    We wish to duduce that P(k+1) is true, that is

    1+5+5^2+...+5^k+1=1/4(5^K+2 -1)

    Now

    1+5+5^2+...+5^k+1=(1+5+5^2+...+5^k)+(5^k+1) (not sure about this line)

    = (cant work out what goes here)

    = 1/4(5^k+2 -1)

    Hence P(K) implies P(k+1), for k=1,2....

    Hence by mathematical induction, p (n) is true, for n=1,2,....
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  6. #6
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    Quote Originally Posted by Arron View Post
    This is how far I have got,

    P(1) is true, since

    1+5^1=6 and 1/4(5^1+1 -1) =6

    Now let k>1, and assume that P(k) is true, that is

    1+5+5^2+...+5^K = 1/4(5^k+1 -).

    We wish to duduce that P(k+1) is true, that is

    1+5+5^2+...+5^k+1=1/4(5^K+2 -1)

    Now

    1+5+5^2+...+5^k+1=(1+5+5^2+...+5^k)+(5^k+1) (not sure about this line)

    = (cant work out what goes here)

    = 1/4(5^k+2 -1)

    Hence P(K) implies P(k+1), for k=1,2....

    Hence by mathematical induction, p (n) is true, for n=1,2,....
    So your induction hypothesis states

    \displaystyle \sum_{n=0}^{k}5^n=\frac{1}{4}\left( 5^{k+1}-1\right)

    Now you wish to show k+1

    \displaystyle \sum_{n=0}^{k+1}5^n= 5^{k+1}+\sum_{n=0}^{k}5^n

    Now use the induction hypothesis to get

    \displaystyle 5^{k+1}+\frac{1}{4}\left( 5^{k+1}-1\right)=\frac{1}{4}\left( 4\cdot 5^{k+1}+5^{k+1}-1\right)=\frac{1}{2}\left( 5\cdot 5^{k+1}-1\right)=\frac{1}{4}\left( 5^{k+2}-1\right)
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  7. #7
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    Quote Originally Posted by Arron View Post
    This is how far I have got,

    P(1) is true, since

    1+5^1=6 and 1/4(5^1+1 -1) =6

    Now let k>1, and assume that P(k) is true, that is

    1+5+5^2+...+5^K = 1/4(5^(k+1) -1)

    We wish to duduce that P(k+1) is true, that is

    1+5+5^2+...+5^(k+1)=1/4(5^(k+2) -1)

    Now

    1+5+5^2+...+5^(k+1)=(1+5+5^2+...+5^k)+5^(k+1) (not sure about this line) It's fine now

    = (cant work out what goes here)

    = 1/4(5^(k+2) -1)

    Hence P(K) implies P(k+1), for k=1,2....

    Hence by mathematical induction, p (n) is true, for n=1,2,....

    P(k)

    5^0+5^1+5^2+....+5^k=\frac{1}{4}\left(5^{k+1}-1\right)

    P(k+1)

    5^0+5^1+5^2+.....+5^{k+1}=\frac{1}{4}\left(5^{k+2}-1\right)

    You should be trying to show that "if" P(k) is true, "then" P(k+1) will also be true.

    Proof

    If P(k) is true, then

    5^0+5^1+....+5^k=\frac{1}{4}\left(5^{k+1}-1\right)

    and so

    5^0+5^1+....+5^k+5^{k+1}=\frac{1}{4}\left(5^{k+1}-1\right)+5^{k+1}

    =\frac{1}{4}\left(5^{k+1}+(4)5^{k+1}-1\right)=\frac{1}{4}\left((5)5^{k+1}-1\right)

    =\frac{1}{4}\left(5^{k+2}-1\right)
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