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# [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

題目是要用fo(x)=1/(1-x)

找f4(x)=.....

下去找.......

### 1 Answer

- 1 decade agoFavorite 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): 我