Question

求解详细过程

Answer 1

a1=2 ,a2=4
a(n+1)+2a(n-1)=3an
a(n+1)-an=2[an-a(n-1)]
[a(n+1)-an]/[an-a(n-1)]=2
{a(n+1)-an}是等比数列
2) a(n+1)-an=(a2-a1)*2^(n-1)=(4-2)*2^(n-1)=2^n
a(n+1)-an=2^n
累加得
a(n+1)-a1=(a2-a1)+(a3-a2)+....+a(n+1)-an=2+2^2+2^3+...+2^n
a(n+1)-a1=2(1-2^n)/(1-2)
a(n+1)=a1+2(1-2^n)/(1-2)=2+2^(n+1)-2
a(n+1)=2^(n+1)
an=2^n
3) Sn=1/2+2/2^2+3/2^3+...+n/2^n
1/2*Sn=1/2^2+2/2^3+3/2^4+...+(n-1)/2^n+n/2^(n+1)
两式相减得
1/2*Sn=1/2+1/2^2+2/2^3+...+n/2^n-n/2^(n+1)
1/2*Sn={1/2[1-1/2^n]/(1-1/2) } - n/2^(n+1)
1/2*Sn=(1-1/2^n) - n/2^(n+1)
1/2*Sn=1-[(n+2)/2^(n+1)]
Sn=2-[(n+2)/2^n]