Question

已知等差数列{an}的首项a1=a,公差d=2,前n项和为Sn,且S1,S2,S4成等比数列

(Ⅰ) 求数列{an}的通项公式；
(2)设bn=2^n*an,求数列bn的前n项和Tn

Answer 1

根据等差数列前n项和公式，Sn=n*a1 + d * n(n-1)/2=na+n(n-1)
S1=a1=a，S2=2a+2，S4=4a+12
S1,S2,S4成等比数列，则S2²=S1 * S4，即(2a+2)²=a(4a+12)，解方程得，a=1

所以通项公式为an=a1+(n-1)d=1+2(n-1)=2n-1

bn=2^n * (2n-1)
Tn=b1+b2+...+bn = 2*1 + 2² * 3 + 2³ * 5+...+2^n (2n-1)
2Tn=2b1+...+2bn= 2² * 1 + 2³ * 3+...+2^n (2n-3) + 2^(n+1)(2n-1) 【对齐了，运用错位相减法
Tn-2Tn=2 * 1 + 2² * 2 + 2³ * 2 +...+2^n * 2 - 2^(n+1) * (2n-1)
=2+ [2³ + 2^4 +... + 2^(n+1)]- 2^(n+1) * (2n-1)
=2+ 2³ * [2^(n-1) - 1] - 2^(n+1) * (2n-1)
=2 + 2^(n+2) -8 - 2^(n+1) * (2n-1)
=-[6 + 2^(n+1) * (2n-1 - 2) ]
=-[6 + 2^(n+1) * (2n-3)]
所以Tn=6 + 2^(n+1) * (2n-3)