1.
[a(n+1)+an][a(n+1)-2an]=0
数列各项均为正,a(n+1)+an>0,因此只有a(n+1)=2an
a(n+1)/an=2,为定值,数列{an}是以2为公比的等比数列.
a3+2是a2、a4的等差中项,则
2(a3+2)=a2+a4
2(2a2+2)=a2+4a2
a2=4
a1=a2/2=4/2=2
an=2×2^(n-1)=2ⁿ,数列{an}的通项公式为an=2ⁿ
2.
bn=-nan=-n×2ⁿ
Sn=b1+b2+...+bn=-(1×2+2×2²+3×2³+...+n×2ⁿ)
令Bn=1×2+2×2²+3×2³+...+n×2ⁿ
则2Bn=1×2²+2×2³+...+(n-1)×2ⁿ+n×2^(n+1)
Bn-2Bn=-Bn=2+2²+...+2ⁿ-n×2^(n+1)
=2×(2ⁿ-1)/(2-1)-n×2^(n+1)
=(1-n)×2^(n+1) -2
Sn=-Bn=(1-n)×2^(n+1) -2