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1/5*1/6+1/6*1/7+1/7*1/8+1/8*1/9+1/9*1/10=1/10...

=1/5*(1-5/6)+1/6*(1-6/7)+1/7*(1-7/8)+1/8*(1-8/9)+1/9*(1-9/10) =1/5-1/10 =1/10

拆分每个分数。记住等式: 1/(x(x+1))=1/x-1/(x+1) 所以,原等式就可以简化为1/1-1/10=0.9

可以用一个for循环依次累加就可以解决: //参考代码 #include int main(){float a=0;int i,j=1,t;for(i=1;i

2004能被2,3,4,6,12整除,所以可以不考虑1/2,1/3,1/4,1/6,1/12 2004除以5,8,10是有限小数,所以也可以不考虑1/5,1/8,1/10 2004/7=286.285714285714,是一个6位的循环,小数点后第2004位,2005位是42 2004/9=222,666是一个1位的循环...

我刚刚答过这个问题,用裂项 1/5*6=1/5-1/6 1/6*7=1/6-1/7 ...... 原式=1/5-1/6+1/6-1/7+......+1/18-1/19+1/19-1/20 =1/5-1/20 =3/20

解:1/(4×5×6)=1/2[1/(4×5)-1/(5×6)] 1/(5×6×7)=1/2[1/(5×6)- 1/(6×7 )] 1/(6×7×8)=1/2[1/(6×7)- 1/(7×8)] ……… 1/(98×99×100)=1/2[1/(98×99)-1/(99×100)] ∴1/(4×5×6)+ 1/(5×6×7)+ 1/(6×7×8)+…..+ 1/(98×99×100) =1/2[1/(4×5)-1/(5×6)]+ 1/2[1/(5...

自己用计算机算就出来了,要不就自己把分母化成一样的再自己++++不懂就找你们老师去.答案来了1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10 =(1/2+1/3+1/6)+(1/4+1/5+1/10)+1/7+1/8+1/9 =1+11/20+1/8+1/7+1/9 =1+27/40+1/9+1/7 =1+283/360+1/7 =1+2341/2...

找出加减号的规律即可

1/[n(n+1)(n+2)]=1/2*[1/n-2/(n+1)+1/(n+2)], 例如 ∴1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+1/(4*5*6)+1/(5*6*7...

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