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1/1×3+1/3×5+1/5×7+1/7×9+........+1/97×99

1/1*3=1/2(1/1-1/3) 1/3*5=1/2(1/3-1/5) 所以原式=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/99)=49/99

1/1*3+1/3*5+1/5*7+1/7*9+1/9*11+…+1/99*101 =2/1*3+2/3*5+2/5*7+2/7*9+2/9*11+…+2/99*101 ÷2=(1-1/3+1/3-1/5+1/5-1/7+……+1/99-1/101)÷2=1-1/101÷2=100/101÷2=50/101

1/1*3+1/3*5+1/5*7+1/7*9+……+1/97*99=[1-1/3]/2 +[1/3-1/5]/2 +[1/5-1/7]/2 +.+[1/97-1/99]/2 =[1-1/3+1/3-1/5+1/5-1/7++1/97-1/99]/2=[1-1/99]/2=49/99

1/1*3+1/3*5+1/5*7+1/7*9+.+1/95*97+1/97*99=(1/1-1/3)/2+(1/3-1/5)/2+(1/5-1/7)/2++(1/97-1/99)/2=(1/1-1/3+1/3-1/5+1/5--1/99)/2=(1/1-1/99)/2=(98/99)/2=49/99 1/(n*(n+2))=(1/n-1/(n+2))/2

1/(1*3)=(1/2)*(1/1-1/3) 1/(3*5)=(1/2)*(1/3-1/5) 1/(5*7)=(1/2)*(1/5-1/7) 所以原式=(1/2)*(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13) =(1/2)*(1-1/13) =6/13

1/1*3=1/2(1-1/3) 1/3*5=1/2(1/3-1/5)1/7*9=1/2(1/5-1/7)1/1*3+1/3*5+1/5*7+1/7*9++1/49*51=1/2(1-1/3+1/3-1/5+..-1/51)=1/2(1-1/51)=25/51希望能帮助你

50/101

1/1*3+1/3*5+1/5*7+1/7*9+.+1/95*97+1/97*99 =(1/2)*[(1-1/3)+(1/3-1/5)+(1/5-1/7)+(1/7-1/9)+……+(1/95-1/97)+(1/97-1/99)] =(1/2)*(1-1/99)=49/99.

1/(1 * 3) = 1/3 = 1/2 * (1 - 1/3)1/(3 * 5) = 1/15 = 1/2 * (1/3 - 1/5)1/(5 * 7) = 1/35 = 1/2 * (1/5 - 1/7)原式= 1/2 * (1 - 1/3) + 1/2 * (1/3 - 1/5) + 1/2 * (1/5 - 1/7) + + 1/2 * (1/97 - 1/99)= 1/2 * (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + + 1/97 - 1/99)= 1/2 * (1 - 1/99)= 1/2 * 98/99= 49/99

观察数字,你会发现这样的规律:1/1*3=(1-1/3)*0.51/3*5=(1/3-1/5)*0.51/5*7=(1/5-1/9)*0.51/7*9=(1/7-1/9)*0.5那么1/1*3+1/3*5+1/5*7+1/7*9+19*11+..+1/1995*1997+1/1997*1999=((1-1/3)+(1/3-1/5)+(1/5-1/9)++(1/1997-1/1999))*0.5=(1-1/1999)*0.5=999/1999还算可以吧

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