求前n项和1.1/(2X5),1/(5X8),1/(8X11)...2.2＾2/(1X3),4＾2/(3X5),6＾2/(5X7)...(2＾2/(1X3)读做2的平方,分之3.下同)裂项相消法不是要你告诉我用什么方法!

Question

求前n项和
1.1/(2X5),1/(5X8),1/(8X11)...
2.2＾2/(1X3),4＾2/(3X5),6＾2/(5X7)...(2＾2/(1X3)读做2的平方,分之3.下同)
裂项相消法
不是要你告诉我用什么方法!

Answer 1

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)