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Thread: Proof of Induction

  1. #1
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    Proof of Induction

    It is given that$\displaystyle sn=2*2+3*2^2+4*2^2+........ +(n+1)*2^n$

    a) By first considering the values of $\displaystyle \frac{S1}{1}$,$\displaystyle \frac{S2}{2}$,$\displaystyle \frac{S3}{3}$,$\displaystyle \frac{S4}{4}$ and making a conjecture for a formula for $\displaystyle \frac{Sn}{n}$, make a conjecture for a formula for Sn.

    b) Use induction to prove that your conjecture is correct.

    For a) i did $\displaystyle S1=4, S2=6, S3=8$

    so the conjecture for $\displaystyle Sn=2n+2$

    b) $\displaystyle S1=4$, $\displaystyle S1=2*1+2=4$
    so n=1 is true.

    $\displaystyle Sk=2k+2$
    $\displaystyle Sk+1=2(k+1)+2 = 2k+4$

    what do i do next? or how to i prove $\displaystyle Sk+1=2k+4$

    thanks.
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  2. #2
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    Quote Originally Posted by BabyMilo View Post
    It is given that$\displaystyle sn=2*2+3*2^2+4*2^2+........ +(n+1)*2^n$

    a) By first considering the values of $\displaystyle \frac{S1}{1}$,$\displaystyle \frac{S2}{2}$,$\displaystyle \frac{S3}{3}$,$\displaystyle \frac{S4}{4}$ and making a conjecture for a formula for $\displaystyle \frac{Sn}{n}$, make a conjecture for a formula for Sn.

    b) Use induction to prove that your conjecture is correct.

    For a) i did $\displaystyle S1=4, S2=6, S3=8$
    No, S2= 2(2)+ 3(4)= 4+ 12= 16, not 6 and S3= 2(2)+ 3(4)+ 4(8)= 4+ 12+ 32= 48, not 8.

    Why not do what you were told to do- look at Sn/n first?

    S1/1= 4, S2/2= 8, S3/3= 48/3= 16, ...

    so the conjecture for $\displaystyle Sn=2n+2$

    b) $\displaystyle S1=4$, $\displaystyle S1=2*1+2=4$
    so n=1 is true.

    $\displaystyle Sk=2k+2$
    $\displaystyle Sk+1=2(k+1)+2 = 2k+4$

    what do i do next? or how to i prove $\displaystyle Sk+1=2k+4$

    thanks.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    No, S2= 2(2)+ 3(4)= 4+ 12= 16, not 6 and S3= 2(2)+ 3(4)+ 4(8)= 4+ 12+ 32= 48, not 8.

    Why not do what you were told to do- look at Sn/n first?

    S1/1= 4, S2/2= 8, S3/3= 48/3= 16, ...
    so i have $\displaystyle \frac{Sn}{n} = 2^{n+1}$

    $\displaystyle \rightarrow Sn=2^{n+1}*n$
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  4. #4
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    Quote Originally Posted by BabyMilo View Post
    so i have $\displaystyle \frac{Sn}{n} = 2^{n+1}$

    $\displaystyle \rightarrow Sn=2^{n+1}*n$
    now im stuck on

    $\displaystyle 2^{k+2}+2(2^{k+1}*k) = 2^{k+2}*(k+1)$
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  5. #5
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    Quote Originally Posted by BabyMilo View Post
    now im stuck on

    $\displaystyle 2^{k+2}+2(2^{k+1}*k) = 2^{k+2}*(k+1)$
    Hi BabyMilo,

    P(k)

    $\displaystyle S_k=k2^{k+1}$

    then we try to express P(k+1) in terms of P(k) to find out if there is a cause

    P(k+1)

    $\displaystyle S_{k+1}=(k+1)2^{k+2}$

    Let's see if this is what we get when we add the (k+1)th term to $\displaystyle S_k$

    $\displaystyle S_k+T_{k+1}=k2^{k+1}+(k+2)2^{k+1}$

    since the (k+1)th term is $\displaystyle (k+1+1)2^{k+1}$

    $\displaystyle S_k+T_{k+1}=2^{k+1}(k+k+2)=2(k+1)2^{k+1}=(k+1)2^{k +2}$

    As $\displaystyle S_k$ being valid will cause $\displaystyle S_{k+1}$ to be valid,

    there is a cause and effect relationship, so you only need test the first one.
    Last edited by Archie Meade; Apr 23rd 2010 at 02:20 AM. Reason: typo
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    Hi BabyMilo,

    P(k)

    $\displaystyle S_k=k2^{k+1}$

    then we try to express P(k+1) in terms of P(k) to find out if there is a cause

    P(k+1)

    $\displaystyle S_{k+1}=(k+1)2^{k+2}$

    Let's see if this is what we get when we add the (k+1)th term to $\displaystyle S_k$

    $\displaystyle S_k+T_{k+1}=k2^{k+1}+(k+2)2^{k+1}$

    since the (k+1)th term is $\displaystyle (k+1)2^{k+1+1}$

    $\displaystyle S_k+T_{k+1}=2^{k+1}(k+k+2)=2(k+1)2^{k+1}=(k+1)2^{k +2}$

    As $\displaystyle S_k$ being valid will cause $\displaystyle S_{k+1}$ to be valid,

    there is a cause and effect relationship, so you only need test the first one.
    Can you show me how you got from to $\displaystyle S_k+T_{k+1}=2^{k+1}(k+k+2)$
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  7. #7
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    Quote Originally Posted by BabyMilo View Post
    Can you show me how you got from to $\displaystyle S_k+T_{k+1}=2^{k+1}(k+k+2)$
    Yes,

    if we take $\displaystyle 2^{k+1}$ as the common factor of both terms

    $\displaystyle 2^{k+1}[k]+2^{k+1}(k+2)=2^{k+1}\left([k]+(k+2)\right)$
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