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