Question

已知数列an中，a1=1,a2=2,an+1-3an+2an-1=0,求an.

Answer 1

0=a(n+1)-3a(n)+2a(n-1)
a(n+1)-a(n)=2[a(n)-a(n-1)]
{a(n+1)-a(n)}是首项为a(2)-a(1)=1,公比为2的等比数列.
a(n+1)-a(n)=2^(n-1)
a(n+1)-2^n = a(n)- 2^(n-1)=...=a(1)-2^(1-1)=0
a(n)=2^(n-1)

Answer 2

an+1-3an+2an-1=0,an+1-an=2(an-an-1),(an+1-an)/(an-an-1)=2,
S(n+1)=a1+a2+┄┄+a(n+1)，Sn=a1+a2+┄┄+an,S(n+1)-Sn=a1+(a2-a1)+(a3-a2)+┄┄+(an-a(n-1))+(a(n+1)-an), a(n+1)=1+2^n-1=2^n, an=2^(n-1)