kkk asked in 科學數學 · 1 decade ago

[數學]離散一題急

Using the summation operator method to find

(1)∑[i=1,n]i^4

(2)∑[i=1,n]i^5

Update:

題目是要用fo(x)=1/(1-x)

找f4(x)=.....

下去找.......

1 Answer

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  • 1 decade ago
    Favorite Answer

    等等會用到的引理:

    (A)Σ[i=1,n]i = n(n+1)/2

    (B)Σ[i=1,n]i^2 = n(n+1)(2n+1)/6

    (C)Σ[i=1,n]i^3 = [n(n+1)/2]^2

    ======================================================

    (1)

    _Σ[i=1,n]i^4

    =(n+1){Σ[i=1,n]i^3}-{Σ[i=1,n]Σ[j=1,i]j^3}

    =(n+1){[n(n+1)/2]^2}-{Σ[i=1,n] [i(i+1)/2]^2}

    =(1/4)(n^2)[(n+1)^3]-(1/4){Σ[i=1,n] (i^2)[(i+1)^2]}

    =(1/4)(n^2)[(n+1)^3]-(1/4){Σ[i=1,n] (i^4)+2(i^3)+(i^2)}

    =(1/4)(n^2)[(n+1)^3]-(1/4){Σ[i=1,n] (i^4)}-(1/2){Σ[i=1,n] (i^3)}-(1/4){Σ[i=1,n] (i^2)}

    =(1/4)(n^2)[(n+1)^3]-(1/2){[n(n+1)/2]^2}-(1/4)(1/6)n(n+1)(2n+1)

    -(1/4){Σ[i=1,n] (i^4)}-------(#)

    (#)式移項後:

    _(5/4){Σ[i=1,n] (i^4)}

    =(1/4)(n^2)[(n+1)^3]-(1/2){[n(n+1)/2]^2}-(1/4)(1/6)n(n+1)(2n+1)

    =........通分化簡後........

    =(1/24)n(n+1)(2n+1)[(3n^2)+3n-1]-------(*)

    (*)式兩端同乘(4/5)後:

    _Σ[i=1,n] (i^4)

    =(1/30)n(n+1)(2n+1)[(3n^2)+3n-1]

    ======================================================

    這題會用到(1)的結果

    (2)

    _Σ[i=1,n]i^5

    =(n+1){Σ[i=1,n]i^4}-{Σ[i=1,n]Σ[j=1,i]j^4}

    =(n+1){(1/30)n(n+1)(2n+1)[(3n^2)+3n-1]}-{Σ[i=1,n](1/30)i(i+1)(2i+1)[(3i^2)+3i-1]}

    =(1/30)n[(n+1)^2](2n+1)[(3n^2)+3n-1]-(1/30){Σ[i=1,n]i(i+1)(2i+1)[(3i^2)+3i-1]}

    =(1/30)n[(n+1)^2](2n+1)[(3n^2)+3n-1]-(1/30){Σ[i=1,n]6(i^5)+15(i^4)+10(i^3)-i}

    =(1/30)n[(n+1)^2](2n+1)[(3n^2)+3n-1]-(1/5){Σ[i=1,n](i^5)}

    +(1/2){Σ[i=1,n](i^4)}-(1/3){Σ[i=1,n](i^3)}+(1/30)Σ[i=1,n]i

    =(1/30)n[(n+1)^2](2n+1)[(3n^2)+3n-1]-(1/2)(1/30){n(n+1)(2n+1)[(3n^2)+3n-1]}-(1/3){[n(n+1)/2]^2}-(1/30){n(n+1)/2}

    -(1/5){Σ[i=1,n](i^5)}------(@)

    (@)式移項後:

    _(6/5){Σ[i=1,n]i^5}

    =(1/30)n[(n+1)^2](2n+1)[(3n^2)+3n-1]-(1/2)(1/30){n(n+1)(2n+1)[(3n^2)+3n-1]}-(1/3){[n(n+1)/2]^2}-(1/30){n(n+1)/2}

    =........通分化簡後........

    =(1/60)n(n+1){[(2n+1)^2][3(n^2)+3n-1]-(5(n^2)+5n-1)}-----($)

    ($)式兩端同乘(5/6)後:

    _Σ[i=1,n]i^5

    =(1/72)n(n+1){[(2n+1)^2][3(n^2)+3n-1]-(5(n^2)+5n-1)}

    2008-11-19 01:19:08 補充:

    題目不是要求"summation operator method"?

    Source(s):
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