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  1. #1
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    sums .

    Deduce the sum of the series .

    (1) $\displaystyle 1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2+(2n+1)^2$


    (2) $\displaystyle 25^2-26^2+27^2-28^2+...+49^2-50^2$


    THanks .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Deduce the sum of the series .

    (1) $\displaystyle 1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2+(2n+1)^2$



    $\displaystyle a^2 -b^2 =(a+b)(a-b)$

    Hence
    $\displaystyle 1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2+(2n+1)^2$

    $\displaystyle = (1+2)(1-2) +~(2+3)(2-3) + ~......... ( ~( 2n-1) + 2n)~) ( ~( 2n-1)- 2n)~) + (2n+1)^2$

    $\displaystyle
    = -1[ 3 +7 +...........(4n-1) ] + (2n+1)^2
    $

    $\displaystyle
    =-1[\frac{n}{2}(2\times 3 + (n-1)4)] + (2n+1)^2
    $............................Using AP

    $\displaystyle
    = -1[ n(2n+1) ] (2n+1)^2
    $

    $\displaystyle = -1[2n^2 +n] + 4n^2 + 1 +4n$

    $\displaystyle = 2n^2 + 3n +1 $

    $\displaystyle = (2n+1)(n+1)$



    (2)
    Now for this try it the same way

    $\displaystyle
    = -1[ 51 +55 +...........(4\times 25 -1) ]
    $

    $\displaystyle
    =-1[\frac{25}{2}(2\times 51 + (13-1)4)]
    $

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