s(n-1)=a(n),
s(n)=a(n+1)=s(n+1)-s(n),
s(n+1)=2s(n).
{s(n)}是首项为s(1)=a(1)=5,公比为2的等比数列.
s(n)=5*2^(n-1),
a(n+1)=s(n)=5*2^(n-1),
a(1)=5,
a(n)=5*2^(n-2),n=2,3,...
a(2)=5,a(3)=5*2=10,a(4)=5*4=20,a(5)=5*8=40.
a(1)=5,
n>=2时,a(n)=5*2^(n-2).
归纳法
a(1)=5,
a(2)=s(2-1)=s(1)=5=5*2^(2-2),满足．
设a(k)=5*2^(k-2),k>2,
则s(k-1)=a(k)=5*2^(k-2),
s(k)=s(k-1)+a(k)=5*2^(k-2)+5*2^(k-2)=5*2^(k-1),
a(k+1)=s(k)=5*2^(k-1)=5*2^(k+1-2),满足．
由归纳法知,a(1)=5,n>=2时,a(n)=5*2^(n-2)成立．