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F. 2 Formulas
發問:
1)The sum(S) of the squares of the first n natural numbers 1^2+2^2+3^2+4^2....+n^2 can be calculated by the formular S =( n (n+1) (2n+1) )/6 Using the formular, find the values of 16^2+17^2+18^2+...30^22)Make the letter in the bracket be... 顯示更多 1)The sum(S) of the squares of the first n natural numbers 1^2+2^2+3^2+4^2....+n^2 can be calculated by the formular S =( n (n+1) (2n+1) )/6 Using the formular, find the values of 16^2+17^2+18^2+...30^2 2)Make the letter in the bracket be the subject of the formular 1/y =(x+1/y+1)+ 1/z [x] 更新: Pls answer it with steps ! 更新 2: sklin1688 no.1 is correct,but no.2's ans wer is x=y-yz-z/z-yz-y 更新 3: Sorry, no.2's question is 1/y=(x-1/x+1)+1/z SorrySorrySorrySorrySorrySorrySorrySorrySorry*1000000000000000000000000000
1)The sum(S) of the squares of the first n natural numbers 1^2+2^2+3^2+4^2....+n^2 can be calculated by the formula S =( n (n+1) (2n+1) )/6 Using the formula, find the values of 16^2+17^2+18^2+...30^2 A. 16^2+17^2+18^2+...30^2 = (1^2+2^2+3^2+4^2...+30^2)-(1^2+2^2+3^2+4^2...+15^2) = {[30(30+1)(60+1)]/6}-{[15(15+1)(30+1)]/6} = (56730/6)-(7440/6) = 9455-1240 = 8215 2) Make the letter in the bracket be the subject of the formula 1/y =(x+1/y+1)+ 1/z [x] A. 1/y =(x+1/y+1)+ 1/z [x] z(y+1) = (x+1)zy+y(y+1) (y+1)(z-y) = yz(x+1) [(y+1)(z-y)]/yz = x+1 x = {[(y+1)(z-y)]/yz}-1 2010-01-10 17:55:33 補充： 2. 1/y=(x-1/x+1)+1/z [x] A. 1/y=(x-1/x+1)+1/z (1/y)(x+1)yz = (x-1/x+1)(x+1)yz+(1/z)(x+1)yz z(x+1) = yz(x-1)+y(x+1) xz+z = xyz-yz+xy+y xz-xyz-xy = y-yz-z x(z-yz-y) = y-yz-z x = (y-yz-z)/(z-yz-y)
其他解答:
F. 2 Formulas
發問:
1)The sum(S) of the squares of the first n natural numbers 1^2+2^2+3^2+4^2....+n^2 can be calculated by the formular S =( n (n+1) (2n+1) )/6 Using the formular, find the values of 16^2+17^2+18^2+...30^22)Make the letter in the bracket be... 顯示更多 1)The sum(S) of the squares of the first n natural numbers 1^2+2^2+3^2+4^2....+n^2 can be calculated by the formular S =( n (n+1) (2n+1) )/6 Using the formular, find the values of 16^2+17^2+18^2+...30^2 2)Make the letter in the bracket be the subject of the formular 1/y =(x+1/y+1)+ 1/z [x] 更新: Pls answer it with steps ! 更新 2: sklin1688 no.1 is correct,but no.2's ans wer is x=y-yz-z/z-yz-y 更新 3: Sorry, no.2's question is 1/y=(x-1/x+1)+1/z SorrySorrySorrySorrySorrySorrySorrySorrySorry*1000000000000000000000000000
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最佳解答:1)The sum(S) of the squares of the first n natural numbers 1^2+2^2+3^2+4^2....+n^2 can be calculated by the formula S =( n (n+1) (2n+1) )/6 Using the formula, find the values of 16^2+17^2+18^2+...30^2 A. 16^2+17^2+18^2+...30^2 = (1^2+2^2+3^2+4^2...+30^2)-(1^2+2^2+3^2+4^2...+15^2) = {[30(30+1)(60+1)]/6}-{[15(15+1)(30+1)]/6} = (56730/6)-(7440/6) = 9455-1240 = 8215 2) Make the letter in the bracket be the subject of the formula 1/y =(x+1/y+1)+ 1/z [x] A. 1/y =(x+1/y+1)+ 1/z [x] z(y+1) = (x+1)zy+y(y+1) (y+1)(z-y) = yz(x+1) [(y+1)(z-y)]/yz = x+1 x = {[(y+1)(z-y)]/yz}-1 2010-01-10 17:55:33 補充： 2. 1/y=(x-1/x+1)+1/z [x] A. 1/y=(x-1/x+1)+1/z (1/y)(x+1)yz = (x-1/x+1)(x+1)yz+(1/z)(x+1)yz z(x+1) = yz(x-1)+y(x+1) xz+z = xyz-yz+xy+y xz-xyz-xy = y-yz-z x(z-yz-y) = y-yz-z x = (y-yz-z)/(z-yz-y)
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