2 |
3 |
∴n≥2时,an=Sn−Sn−1=
2 |
3 |
2 |
3 |
∴n≥2时,
an |
an−1 |
∴数列an是首项为a1=3,公比为q=-2的等比数列,
∴an=3•(-2)n-1,n∈N*
(Ⅱ)由(Ⅰ)知,n|an|=3n•2n-1.
∴Tn=3(1+2•21+3•22+4•23+…+n•2n-1)
2Tn=3(1•21+2•22+3•23+…+(n-1)•2n-1+n•2n)
∴-Tn=3(1+2+22+23+…+2n-1-n•2n)
∴−Tn=3[
1−2n |
1−2 |
∴Tn=3+3n•2n-3•2n