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高中1/(1+9)+1/(1+10)+1/(1+11)+1/(1+12)+....+1/(...

1/(1+2+3+……+9)+1/(1+2+3+……+10)+1/(1+2+3+……+11)+1/(1+2+3+……+12)+....+1/(1+2+3+……+100) =2/(9×10)+2/(10×11)+2/(11×12)+2/(12×13)+....+2/(100×101) =2×(1/9-1/10+1/10-1/11+1/11-1/12+1/12-1/13+……+1/100-1/101) =2×(1/9-1/101) =2×92/9...

(1/8+1/9+1/10+1/11)×(1/9+1/10+1/11+1/12)-(1/8+1/9+1/10+1/11+1/12)×(1/9+1/10+1/11) =(1/8+1/9+1/10+1/11)×[(1/9+1/10+1/11)+1/12] - [(1/8+1/9+1/10+1/11)+1/12]×(1/9+1/10+1/11) =(1/8+1/9+1/10+1/11)×(1/9+1/10+1/11)+(1/8+1/9+1/10+1/11)...

设共同都有的一个为a 设1/9+1/10+1/11=a ∴原式可化为 (1/8+a)(a+1/12)-(1/8+a+1/12)a =1/8a+1/96+a²+1/12a-1/8a-a²-1/12a =1/96 关键是整体思想

令a=1/9+1/10+1/11 原式=(1/8+a)(a+1/12)-(1/8+a+1/12)a =1/8(a+1/12)+a(a+1/12)-1/8a-a(a+1/12) =1/8(a+1/12)-1/8a =1/8a+1/96-1/8a =1/96

首先提取公因式,归纳一下,就可以化成1/15*(1+2+.....+14)+1/14*(1+2+....+13)+........1/3*(1+2)+1/2 然后根据括号内可以用等差求和的方法每个单项式归纳为1/n*n/2*(n-1)=(n-1)/2 所以将N分别取15到2 提取公因式1/2 最后变成(1+3+4+......

设1/9+1/10+1/11=a ∴(1/8+1/9+1/10+1/11)*(1/9+1/10+1/11+1/12)-(1/8+1/9+1/10+1/11+1/12)*(1/9+1/10+1/11) =(1/8+a)(a+1/12)-(1/8+a+1/12)a =1/96+a/8+a/12+a²-a/8-a²-a/12 =1/96

1÷(1/15+1/10)=6 1÷(1/15+1/10)=6 1÷(1/20+1/30)=12

1/(7*8)+1/(8*9)+1/(9*10)+1/(10*11)+1/(11*12) =(1/7-1/8)+(1/8+1/9)+...+(1/11-1/12) =1/7 -1/12 = 5/84

1/12+1/20+1/30+…+1/90 =(1/3-1/4)+(1/4-1/5)+..+(1/9-1/10) =1/3-1/4+1/4-1/5+1/5-1/6+..+1/9-1/10 =1/3-1/10 =10/30-3/30 =7/30; 您好,很高兴为您解答,skyhunter002为您答疑解惑 如果本题有什么不明白可以追问,如果满意记得采纳 如果有其他...

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