发布网友 发布时间:2024-10-24 09:58
共1个回答
热心网友 时间:2024-10-29 22:45
1/10×11+1/11×12+1/12×13+.+1/20×21=? 1/10×11+1/11×12+1/12×13+...+1/20×21
=(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+...+(1/20-1/21)
=1/10-1/11+1/11-1/12+1/12-1/13+...+1/20-1/21
=1/10-1/21
=11/210
裂项相消法。
1/10*11+1/11*12+1/12*13+...+1/19*20
=(1/10-1/11)+(1/11-1/12)+...+(1/19-1/20)
=1/10+(1/11-1/11)+(1/12-1/12)...+(1/19-1/19)-1/20
=1/10-1/20
=1/20
1/(10*11)=(11-10)/(10*11)=11/(10*11)-10/(10*11)=1/10-1/11。
之后几项同样处理。
那么,原式=1/10-1/20=1/20 (中间几项都抵消了,只剩一头一尾了)
1/10*11+1/11*12+1/12*13+.1/19*20+1/20*21=?
可以利用1/n-1/(n+1)=1/[n*(n+1)]这个规律来转换题目,转换后除了最前面和最后面一项,其余项之和为0,过程如下:
1/10*11+1/11*12+1/12*13+......1/19*20+1/20*21
=(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+……(1/19-1/20)+(1/20-1/21)
=1/10-(1/11+1/11-1/12+1/12-1/13+......+1/19-1/20+1/20)-1/21
=1/10-0+1/21
=1/10*21
=1/210
1/(10*11)=1/10-1/11
1/10*11+1/11*12+1/12*13........1/19*20
=(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+...+(1/18-1/19)+(1/19-1/20)
=1/10-1/20
=1/20
由于1/10*11=1/10-1/11
1/11*12=1/11-1/12
1/12*13=1/12-1/13
..................
..................
所以原式==(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+......+(1/19-1/20)
=1/10-1/11+1/11-1/12+1/12-1/13+......+1/19-1/20
=1/10-1/20
=1/20
如果题目没写错
-1/10*11+1/11*12+1/12*13+1/13*14+···1/19*20
= -1/10+1/11+1/11-1/12++···+1/19-1/20
=-1/10+1/11+1/11-1/20
=2/11-3/20
如果是
1/10*11+1/11*12+1/12*13+1/13*14+···1/19*20
=1/10-1/20=1/20 中间相消去
=(1/10-1/11)+(1/11-1/12)+..+(1/99-1/100)
=1/10-1/100
=9/100
原式=(1/10-1/11)+(1/11-1/12)+......+(1/19-1/20)=1/10-1/20=1/20