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.