1.
n≥2時,a(n-1)+1=2an
2an-2=a(n-1)-1
(an -1)/[a(n-1)-1]=1/2,為定值.
a1-1=1/2-1=-1/2,數(shù)列{an -1}是以-1/2為首項(xiàng),1/2為公比的等比數(shù)列.
an -1=(-1/2)(1/2)^(n-1)=-1/2ⁿ
an=1- 1/2ⁿ
數(shù)列{an}的通項(xiàng)公式為an=1- 1/2ⁿ
2.
n=1時,2b1=1×2=2b1=1
n≥2時,
2b1+2²b2+...+2ⁿbn=n×2ⁿ (1)
2b1+2²b2+...+2^(n-1)b(n-1)=(n-1)×2^(n-1)(2)
(1)-(2)
2ⁿbn=n×2ⁿ-(n-1)×2^(n-1)=(n+1)×2^(n-1)
bn=(n+1)/2
n=1時,b1=(1+1)/2=1,同樣滿足通項(xiàng)公式
數(shù)列{bn}的通項(xiàng)公式為bn=(n+1)/2
3.
cn=2anbn=2(1-1/2ⁿ)(n+1)/2=(n+1) -(n+1)/2ⁿ
Tn=c1+c2+...+cn=[2+3+...+(n+1)]-[2/2+3/2²+4/2³+...+(n+1)/2ⁿ]
=[1+2+...+(n+1)-1]-[2/2+3/2²+4/2³+...+(n+1)/2ⁿ]
=(n+1)(n+2)/2 -1 -[2/2+3/2²+4/2³+...+(n+1)/2ⁿ]
令Cn=2/2+3/2²+4/2³+...+(n+1)/2ⁿ
則Cn /2=2/2²+3/2³+...+n/2ⁿ +(n+1)/2^(n+1)
Cn -Cn /2=Cn /2=1+1/2²+1/2³+...+1/2ⁿ -(n+1)/2^(n+1)
=1+1/2+1/2²+...+1/2ⁿ -(n+1)/2^(n+1) -1/2
=1×[1-1/2^(n+1)]/(1-1/2) -(n+1)/2^(n+1) -1/2
= -(n+3)/2^(n+1) +3/2
Cn=-(n+3)/2ⁿ +3
Tn=(n+1)(n+2)/2 -1 -Cn
=(n+1)(n+2)/2 -1 +(n+3)/2ⁿ -3
=(n+1)(n+2)/2 +(n+3)/2ⁿ -4