1除以(n+3)(n+4)+1除以(n+4)(n+5)+、、、1除以(n+10)(n+11)=?

问题描述:

1除以(n+3)(n+4)+1除以(n+4)(n+5)+、、、1除以(n+10)(n+11)=?

原式=1/(n+3)-1/(n+4)+1/(n+4)-……+1/(n+10)-1/(n+11)
=1/(n+3)-1/(n+11)
=8/(n+3)(n+11)

设n+3=a
原式=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)+……+1/(a+7)-1/(a+8)
=1/a-1/(a+8)=1/(n+3)-1/(n+11)

1除以(n+3)(n+4)
等于1/(n+3) - 1/(n+4)
故把上式可化简为
1/(n+3) -1/(n+11)

1除以(n+3)(n+4)+1除以(n+4)(n+5)+、、、1除以(n+10)(n+11)
=1/(n+3)-1/(n+4)-1/(n+4)-1/(n+5)+……+1/(n+10)-1/(n+11)
=1/(n+3)-1/(n+11)

1/[n(n+1)]=1/n-1/(n+1)套用这个公式1除以(n+3)(n+4)+1除以(n+4)(n+5)+、、、1除以(n+10)(n+11)=1/(n+3)-1/(n+4)+1/(n+4)-1/(n+5)+...+1/(n+10)-1/(n+11)=1/(n+3)-1/(n+11)=(n+11-n-3)/[(n+3)(n+11)=8/(n^2+14n+33...

裂项抵消后是
1/(n+3)-1/(n+11)=8/(n+3)(n+11)