观察下列三组等式,

题目

观察下列三组等式,
①1/(1x2)=1/1-1/2,1/(2x3)=1/2-1/3,1/(3x4)=1/3-1/4；
②2/(1x3)=1/1-1/3,2/(2x4)=1/2-1/4,2/(3x5)=1/3-1/5；
③3/(1x4)=1/1-1/4,3/(2x5)=1/2-1/5,3/(3x6)=1/3-1/6.
问：计算1/(3x6)+1/(6x9)+1/(9x12)+.+1/3nx(3n+3)
问：计算1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+.+1/(9^2-1)+1/(10^2-1)

答案

1)=1/3*(1/3-1/6+1/6-1/9+.-1/3n+3)
=1/3*(1/3-1/3n+3)
=n/(9n+9)
2)=1/[(2+1)(2-1)]+.+1/[(10+1)(10-1)]
=1/2*(1-1/3+.1/9-1/11)
=1/2*(1-1/11)
=5/11