数列{an}为等差数列，an＝n，数列{bn}为等比数列，bn＝(1⼀2)^(n－1)，求数列{an*bn}的前n项和Sn

a1=1 ， b1=(1/2)^(1-1)=1
an*bn=n/2^(n-1)
Sn=a1*b1+a2*b2+...+an*bn=1/2^0+2/2^1+3/2^2+...+n/2^(n-1) 一
Sn/2=1/2^1+2/2^2+...+(n-1)/2^(n-1)+n/2^n 二
一减二得：Sn-Sn/2=Sn/2=1/2^0+1/2^1+1/2^2+...+1/2^(n-1)-n/2^n
=(1-1/2^n)/(1-1/2)-n/2^n
=2-2/2^n -n/2^n
=2-(n+2)/2^n
Sn=4-(n+2)/2^(n-1)

a1=1 b1=(1/2)^(1-1)=1
anbn=n/2^(n-1)
Sn=a1b1+a2b2+...+anbn=1/2^0+2/2^1+3/2^2+...+n/2^(n-1)
Sn/2=1/2^1+2/2^2+...+(n-1)/2^(n-1)+n/2^n
Sn-Sn/2=Sn/2=1/2^0+1/2^1+1/2^2+...+1/2^(n-1)-n/2^n
=(1-1/2^n)/(1-1/2)-n/2^n
=2-2/2^n -n/2^n
=2-(n+2)/2^n
Sn=4-(n+2)/2^(n-1)

b1=1
q=b2/b1=(1/2)^1/1=1/2
an*bn=n/2^(n-1)
Sn=1*b1+2*b2+...+n*bn
=(b1+b2+...+bn)+(b2+b3+...+bn)+...[b(n-1)+bn]+bn
=b1*(1-q^n)/(1-q)+b2*[1-q^(n-1)]/(1-q)+...+bn*(1-q^1)/(1-q)
={b1*(1-q^n)+b1*q*[1-q^(n-1)]+...+b1*q^(n-1)*(1-q^1)}/(1-q)
=[1-q^n+q-q^n+...+q^(n-1)-q^n]*b1/(1-q)
=[1+q+q^2+...+q^(n-1)-n*q^n]*b1/(1-q)
={[1-q^(n-1)]/(1-q)-n*q^n]}*b1/(1-q)
={[1-(1/2)^(n-1)]/(1-1/2)-n*(1/2)^n]}*1/(1-1/2)
=4-(1/2)^(n-3)]-n*(1/2)^(n-1)