高一数学简单的数列问题

1/(2X5)+1/(5X8)+1/(8X11)+...+1/((3n-1)*(3n+2))
=1/3*(1/2-1/5+1/5-1/8+1/8-1/11+...+1/(3n-1)-1/(3n+2))
=1/3*(1/2-1/(3n+2))
=n/(6n+4)

2^2/(1X3)+4^2/(3X5)+6^2/(5X7)+...+(2n)^2/((2n-1)*(2n+1))
因为
(2n)^2/((2n-1)*(2n+1)
=((2n)^2-1+1)/((2n)^2-1)
=1+1/((2n)^2-1)
=1+1/2*(1/(2n-1)-1/(2n+1))
所以原式=
1+1/2*(1/1-1/3)+1+1/2*(1/3-1/5)+1+1/2*(1/5-1/7)+...+1+1/2*(1/(2n-1)-1/(2n+1))
=1*n+1/2*(1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1))
=n+1/2*(1-1/(2n+1))
=(2n^2+2n)/(2n+1)

裂项相消法

1.=1/3（1/2-1/5+1/5-/1/8+1/8-1/11.....+1/（3n-1）-1/（3n-2) )=(3n-4)/(18n-12)

(1)观察上式可得：
1/(2X5)=(1/3)(1/2-1/5)
1/(5X8)=(1/3)(1/5-1/8)
1/(8X11)=(1/3)(1/5-1/8)
……
1/[(3n-1)(3n+2)]=(1/3)[1/(3n-1)-1/(3n+2)]
把以上的式子都加起来可得：
Sn=(1/3)[(1/2)-(1/5)+(1/5)-(1/8)+……+1/(3n-1)-1/(3n+2)]
=(1/3)[(1/2)-1/(3n+2)]
=n/(6n+4)

(2)帮不到你了，不过可能是用错位相减～～～