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Math Help - Need help on Mathematical Induction question

  1. #1
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    Need help on Mathematical Induction question

    Someone please help? i'm stuck this is how far i got.

    For statement

    1*3 + 2*4 + 3-5 ... + (n-1)*(n+1) = (n(n-1)(2n+5)/6 )for all integers >= 2.

    Base step(B)

    P(2) is true since (2-1)x(2+1) = (2(2-1)(2(2)+5)/6)
    3 = 3
    thus P(2) is true.

    Inductive step (R)

    Assume P(n) is true for some n E N.
    Prove that (n) x (n+2) = [(n+1)(n)(2n+1 +5)/6] is true. <-- is this correct? it looks wierd for some reason.

    (n-1)x(n+1) + n x (n+2) = (n(n-1)(2n+5)/6 ) + n(n+2)
    = ((n(n-1)(2n+5) + 6[n(n+2)]/6)

    and this is where i got stuck. i durno how to simplify the equation could someone please tell me how to solve this problem?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chogz View Post
    Someone please help? i'm stuck this is how far i got.

    For statement

    1*3 + 2*4 + 3-5 ... + (n-1)*(n+1) = (n(n-1)(2n+5)/6 )for all integers >= 2.

    Base step(B)

    P(2) is true since (2-1)x(2+1) = (2(2-1)(2(2)+5)/6)
    3 = 3
    thus P(2) is true.

    Inductive step (R)

    Assume P(n) is true for some n E N.
    Prove that (n) x (n+2) = [(n+1)(n)(2n+1 +5)/6] is true. <-- is this correct? it looks wierd for some reason.

    (n-1)x(n+1) + n x (n+2) = (n(n-1)(2n+5)/6 ) + n(n+2)
    = ((n(n-1)(2n+5) + 6[n(n+2)]/6)

    and this is where i got stuck. i durno how to simplify the equation could someone please tell me how to solve this problem?
    What you need to do is show that

    (1\cdot3)+(2\cdot4)+(3\cdot5)+\dots+[(n-1)\cdot(n+1)]+[n\cdot(n+2)]=\frac{(n+1)(n)(2n+7)}{6}

    But,

    \underbrace{(1\cdot3)+(2\cdot4)+(3\cdot5)+\dots+[(n-1)\cdot(n+1)]}_{\frac{n(n-1)(2n+5)}{6}}+[n\cdot(n+2)]=\frac{(n+1)(n)(2n+7)}{6}

    So all you need to do now is get the left side to look like the right side:

    \frac{(n-1)(n)(2n+5)}{6}+n(n+2)=\frac{n(n+1)(2n+7)}{6}

    Can you take it from here?

    --Chris
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  3. #3
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    Hi chogz,

    P(n):~~~\sum^n_{r=1}{(r-1)(r+1)}=\frac{1}{6}n(n-1)(2n+5)

    Assume P(k) is true for some k \in \mathbb {N}^+ and hence,

    \sum^k_{r=1}{(r-1)(r+1)}=\frac{1}{6}k(k-1)(2k+5)

    \implies \sum^{k+1}_{r=1}{(r-1)(r+1)}=\frac{1}{6}k(k-1)(2k+5)+k(k+2)

    = \frac{1}{6}k[(k-1)(2k+5)+6(k+2)]

    = \frac{1}{6}k[2k^2+9k+7]

    = \frac{1}{6}k(k+1)(2k+7)

    = \frac{1}{6}(k+1)[(k+1)-1][2(k+1)+5]

    hence P(k)\implies P(k+1) is true and thus P(n) is true \forall n \in \mathbb {Z} and n\geq 2

    Hope this helps.
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  4. #4
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    Thanks alot guys you've helped big time.

    Really appreciate it and once again thank you.
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  5. #5
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    Just to confirm:

    <---



    could you please explain why the k from where i indicated was taken out? in the next line, and to make sure the 6 came from the fraction side rite? so it be something like 6(k+2)/6 ? wouldn't it? if possible could you show the working out how you got to the next line? because i don't really fully understand it.

    sorry i'm not really good with these types of question.
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  6. #6
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    sorry my bad.

    Nevermind, i figured it out why the k was removed; throught simplification.

    thanks again guys it really helped.
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