Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Show that ∑(n+ r-1)(n+r) = (1/3n)(7n^2 – 1)?

Show that ∑(n+ r-1)(n+r) = 1/3n(7n^2 – 1)

Find ∑1/(n+ r-1)(n+r) in terms of n

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  • lou h
    Lv 7
    1 decade ago
    Best Answer

    ∑ _{r=1, n} 1= n

    ∑ _{r=1, n} r= (n+1)·n/2

    ∑ _{r=1, n} r²= n·(n+1)·(2n+1)/6

    ==> ∑ _{r=1, n} (n+ r-1)(n+r) = ∑ _{r=1, n} (n²+2nr +r² -n -r) =

    (n²-n)·∑ _{r=1, n} 1 + (2n-1)· ∑ _{r=1, n} r + ∑ _{r=1, n} r² =

    (n²-n)·n + (2n-1)·(n+1)·n/2 + n·(n+1)·(2n+1)/6 =

    n/6 ·[ 6·(n²-n)+3·(2n-1)·(n+1) +(n+1)·(2n+1)] =

    n/6· [6n²-6n+6n²+3n-3+2n²+3n+1] =

    n/6· [14n²-2] = 1/3· n·(7n²-1)

    ---------------------

    1/[(n+r-1)·(n+r)]= 1/(n+r-1) - 1/(n+r) ==>

    ∑ _{r=1, n} 1/[(n+r-1)·(n+r)]= ∑ _{r=1, n} [1/(n+r-1) - 1/(n+r) ] =

    [ 1/n - 1/(n+1) ] + [ 1/(n+1) - 1/(n+2)] +....+ [1/(2n-1)- 1/(2n)] =

    1/n - 1/(2n) = 1/(2n)

    Saludos

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