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Math Help - summation of progressions

  1. #1
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    summation of progressions

    I have bunch of these kind of tasks at school but i dont know where to start. If someone here could show me how to complete tasks like that so i could do other tasks too.

    Prove that, 1x2+2x3+3x4+...+n(n+1)=n(n+1)(n+2)
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    Last edited by CaptainBlack; December 4th 2011 at 03:53 AM.
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  2. #2
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    Re: Some stuff

    Quote Originally Posted by Trolo View Post
    I have bunch of these kind of tasks at school but i dont know where to start. If someone here could show me how to complete tasks like that so i could do other tasks too.

    Prove that, 1x2+2x3+3x4+...+n(n+1)=N(N+1)(N+2)
    ------------
    3
    Am I correct in assuming that n and N are the same number? If so, use the same symbol please.
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Some stuff

    Do you know the following formulas?
    1) \sum_{i=1}^{n}i=\frac{n(n+1)}{2}
    2) \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}


    Note That: 1 \cdot 2 +2 \cdot 3+3 \cdot 4+...+n(n+1)=\sum_{i=1}^{n}i(i+1)

    =\sum_{i=1}^{n}(i^2+i)=\sum_{i=1}^{n}i^2+\sum_{i=1  }^{n}i
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  4. #4
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    Re: Some stuff

    No, ill search for it. Thanks
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  5. #5
    Member sbhatnagar's Avatar
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    Re: Some stuff

    in case, you don't find it....

    \text{Prove} \sum_{i=1}^{n}i=\frac{n(n+1)}{2}
    \sum_{i=1}^{n}i=\underbrace{1+2+3+4+5+6+7+8+...+n}  _{\text{This is an arithmetic progression with d=1 and a=1.}}

    Therefore: \sum_{i=1}^{n}i=\frac{n}{2}{\left[ 2+(n-1)\right]}=\frac{n(n+1)}{2}

    \text{Prove} \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}
    Consider the identity: (x+1)^3-x^3=3x^2+3x+1

    Putting x=1 &,2 &,3 &, ......(n-1) &, n we get:

    \boxed{\begin{matrix} 2^3 &-1^3&=&3\cdot1^2 & +3\cdot 1 &+1 \\ 3^3 &-2^3&=&3\cdot2^2 & +3\cdot 2 &+1 \\4^3 &-3^3&=&3\cdot3^2 & +3\cdot 3 &+1 \\ ..&..&&..&..&.. \\ ..&..&&..&..&.. \\ n^3&-(n-1)^3&=&3\cdot (n-1)^2&+3\cdot (n-1)&+1 \\ (n+1)^3&-(n)^3&=&3\cdot (n)^2&+3\cdot (n)&+1 \end{matrix}}

    Adding column wise, we obtain:

    (n+1)^3-1^3=3{\left(\sum_{i=1}^{n}i^2 \right)}+3{\left(\sum_{i=1}^{n}i \right)}+n

    \implies (n+1)^3-1^3=3{\left(\sum_{i=1}^{n}i^2 \right)}+3{\left \{\frac{n(n+1)}{2} \right \}}+n

    Arranging things a little bit...

    \implies \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}
    Last edited by sbhatnagar; December 4th 2011 at 03:23 AM.
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