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2+6+12+20+30+42+... how to find the Sum of the first 18 this the the equ of sum18(how does this equ come out???): 1/3(n^3) + n^2 + 2/3n
最佳解答:
First, we notice that this is twice the values of Pascal triangle. Thus the series belongs to a class of number called figurate numbers, i.e. the numbers will make different shapes, or figures. In this particular case, the will make two triangles of different orders. For more information on fiurage numbers, follow the following links: Wolfram, or YoungZones.org . To proceed, we will find a general formula to write down the nth term of the series by the method of differences. This method applies to numbers which can be represented by a polynomial, in which case the differences eventually become zero. For the case in point: Series: 2 6 12 20 30 42 ... 1st diff: 4 6 8 10 12 ... 2nd diff: 2 2 2 2 2 ... 3rd diffL 0 0 0 0 0 ... The general formula for the nth term is: T(n)=T(1) + D1(1)*(n-1)/1! + D2(1)*(n-1)(n-2)/2! + D3(1)*(n-1)(n-2)(n-3)/3! + ... where T(1)=first term, D1(1)=first term of the 1st difference D2(1)=first term of the 2nd difference, etc. Substituting into the case above, nth term =T(n) =2 + 4(n-1)/1! + 2(n-1)(n-2)/2! + 0 =2 + 4(n-1) + (n-1)(n-2) =n(n+1) The sum to nth term is: S(n) =T(1)n/1! + D1(1)n(n-1)/2! + D2(1)n(n-1)(n-2)/3! + ... =2n + 4n(n-1)/2! + 2n(n-1)(n-2)/3! =2n+2n(n-1)+n(n-1)(n-2)/3 =n(n+1)(n+2)/3 (=1/3(n^3) + n^2 + 2/3n) Thus for n=18,x(x+1) T(18)=18*19=342x(x+1) S(18)=18*19*20/3=2280 The derivation of the above formulas can be found on p.320 of: H. S. Hall and S.R. Knight, Higher Algebra, London: Macmillan Co., 1899.
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pre-cal發問:
2+6+12+20+30+42+... how to find the Sum of the first 18 this the the equ of sum18(how does this equ come out???): 1/3(n^3) + n^2 + 2/3n
最佳解答:
First, we notice that this is twice the values of Pascal triangle. Thus the series belongs to a class of number called figurate numbers, i.e. the numbers will make different shapes, or figures. In this particular case, the will make two triangles of different orders. For more information on fiurage numbers, follow the following links: Wolfram, or YoungZones.org . To proceed, we will find a general formula to write down the nth term of the series by the method of differences. This method applies to numbers which can be represented by a polynomial, in which case the differences eventually become zero. For the case in point: Series: 2 6 12 20 30 42 ... 1st diff: 4 6 8 10 12 ... 2nd diff: 2 2 2 2 2 ... 3rd diffL 0 0 0 0 0 ... The general formula for the nth term is: T(n)=T(1) + D1(1)*(n-1)/1! + D2(1)*(n-1)(n-2)/2! + D3(1)*(n-1)(n-2)(n-3)/3! + ... where T(1)=first term, D1(1)=first term of the 1st difference D2(1)=first term of the 2nd difference, etc. Substituting into the case above, nth term =T(n) =2 + 4(n-1)/1! + 2(n-1)(n-2)/2! + 0 =2 + 4(n-1) + (n-1)(n-2) =n(n+1) The sum to nth term is: S(n) =T(1)n/1! + D1(1)n(n-1)/2! + D2(1)n(n-1)(n-2)/3! + ... =2n + 4n(n-1)/2! + 2n(n-1)(n-2)/3! =2n+2n(n-1)+n(n-1)(n-2)/3 =n(n+1)(n+2)/3 (=1/3(n^3) + n^2 + 2/3n) Thus for n=18,x(x+1) T(18)=18*19=342x(x+1) S(18)=18*19*20/3=2280 The derivation of the above formulas can be found on p.320 of: H. S. Hall and S.R. Knight, Higher Algebra, London: Macmillan Co., 1899.
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