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Math Help - prove by induction

  1. #1
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    prove by induction

    plz help me with this

    prove by mathematical induction
    P(n):1+3+6+...........n(n+1)(n+2)/2=n(n+1)(n+2)/6
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  2. #2
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    Quote Originally Posted by chhatrashal View Post
    plz help me with this

    prove by mathematical induction
    P(n):1+3+6+...........n(n+1)(n+2)/2=n(n+1)(n+2)/6
    It's no wonder you are having problems. It isn't true.

    You have either typed it incorrectly or your original problem statement has a typo.

    -Dan

    Edit:
    As I recall
    \sum_{i = 1}^{n}i^2 = \frac{n(n + 1)(n + 2)}{6}

    Is this your problem?
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  3. #3
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    But that n+2 should be 2n+1.
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  4. #4
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    Hello, chhatrashal!

    Yes, there is a typo . . .


    Prove by mathematical induction:
    . . P(n)\!:\;\;1+3+6+ \hdots + {\color{blue}\frac{n(n+1)}{2}} \;=\;\frac{n(n+1)(n+2)}{6}
    This is the sum of triangular numbers . . . and it's true.


    Verify S(1)\!:\;\;1 \:=\:\frac{1\cdot2\cdot3}{6} . . . True.


    Assume S(k)\!:\;\;1 + 3 + 6 + \hdots + \frac{k(k+1)}{2} \;=\;\frac{k(k+1)(k+2)}{6}


    Add \frac{(k+1)(k+2)}{2} to both sides;

    . . \underbrace{1 + 3 + 6 + \hdots + \frac{(k+1)(k+2)}{2}}_{\text{The left side of }S(k+1)} \;=\;\frac{k(k+1)(k+2)}{6} + \frac{(k+1)(k+2)}{2}

    The right side is: . \frac{k(k+1)(k+2)}{6} + \frac{(k+1)(k+2)}{2} \;=\;\frac{k(k+1)(k+2)}{6} + \frac{3(k+1)(k+2)}{6}

    . . \text{Factor: }\;\frac{(k+1)(k+2)}{6}\bigg[k + 3\bigg] \;=\;\underbrace{\frac{(k+1)(k+2)(k+3)}{6}}_{\text  {The right side of }S(k+1)}


    We have established S(k+1) . . . The inductive proof is complete.

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