几题数学题,帮解一下thanks

3个回答

  • 1.原式=2*0*(2*0+1)+2*1*(2*1+1)

    .2*1005*(2*1005+1)

    =2*1*(2*1+1).2*1005*(2*1005+1)

    其通项公式=2*n(2*n+1)

    =2*(2n^2+n)

    =4n^2+2n

    =4*(1^2+2^2.1005)^2+2(1+2+3.1005)

    =4*[1005*1006*2011]/6+1005*1006

    =1356465250

    2.原式=(7-6)/6*7+(8-7)/7*8...+(101-100)/100*101

    =(1/6)-(1/7)+(1/7)-(1/8).+(1/100)-(1/101)

    =(1/6)-(1/101)

    3.原式=(1+1/2)(1+1/3)(1+1/4)..(1+1/99)*(1-1/2)(1-1/3)...(1-1/99)

    =(3/2)(4/3)(5/4)...(100/99)*(1/2)(2/3)*(3/4).(98/99)

    =(100/2)*(1/99)

    =50/99

    谜语

    1.公式

    2.区间