数列{1+(1+2)+(1+2+4)+…+(1+2+…+2n-1)}的前n项和为______.

2个回答

  • 解题思路:先求出1+2+…+2n-1=2n-1,再用分组求和法求出1+(1+2)+(1+2+4)+…+(1+2+…+2n-1)=2n+1-2-n.由此能求出数列{1+(1+2)+(1+2+4)+…+(1+2+…+2n-1)}的前n项和.

    ∵1+2+…+2n-1=

    1−2n

    1−2=2n-1,

    ∴1+(1+2)+(1+2+4)+…+(1+2+…+2n-1

    =(2+22+23+…+2n)-n

    =

    2(1−2n)

    1−2-n

    =2n+1-2-n.

    ∴数列{1+(1+2)+(1+2+4)+…+(1+2+…+2n-1)}的前n项和:

    Sn=(22+23+…+2n+1)-2n-(1+2+3+…+n)

    =

    4(1−2n)

    1−2-2n-

    n(n+1)

    2

    =2n+2−

    n2+5n

    2-4.

    故答案为:2n+2−

    n2+5n

    2-4.

    点评:

    本题考点: 数列的求和.

    考点点评: 本题考查数列的前n项和的求法,是中档题,解题时要认真审题,注意分组求和法的合理运用.