数列{1+(1+2)+(1+2+4)+…+(1+2+…+2n-1)}的前n项和为_.
问题描述:
数列{1+(1+2)+(1+2+4)+…+(1+2+…+2n-1)}的前n项和为______.
答
∵1+2+…+2n-1=1−2n1−2=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...