已知数列{an}满足an+1=2an+3.5^n,a1=6.求an

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!

3分之14乘2的n-2次方乘(1.75的n减1次方-1)