已知数列{an}满足an+1=2an+3.5^n,a1=6.求an
2个回答

a(n+1)-2an=3.5^n,则

a2-2a1=3.5^1

a3-2a2=3.5^2

.

a(n+1)-2an=3.5^n

以上式子相加,得

a(n+1)-a1-Sn=3.5+3.5^2+...+3.5^n=3.5(1-3.5^n)/(1-3.5)=7(3.5^n-1)/5

a(n+1)=S(n+1)-Sn--->a(n+1)-S(n+1)=-Sn

-Sn-a1=7(3.5^n-1)/5--->Sn=-a1-7(3.5^n-1)/5=-6-7(3.5^n-1)/5

an=Sn-S(n-1)=[-6-7(3.5^n-1)/5]-[-6-7(3.5^(n-1)-1)/5]

=[7*3.5^(n-1)-7*3.5^n]/5

=-3.5^n

有点怪异!