求前n项和1.1/(2X5),1/(5X8),1/(8X11)...2.2＾2/(1X3),4＾2/(3X5),6＾2/
4个回答

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)

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