思路：分奇偶项讨论,然后分别相加求和.
a(2n+1)=[3+(-1)^(2n)]a(2n)/2=2a(2n)
a(2n+2)=a(2n+1+1)=[3+(-1)^(2n+1)]a(2n+1)/2=a(2n+1)=2a(2n)
{a(2n)}是首项为a(2)=3,公比为2的等比数列.
a(2n)=3*2^(n-1)
a(2n+1)=2a(2n)=3*2^n
a(1)+a(2)+...+a(100)=
=a(1)+a(3)+a(5)+...+a(99) + a(2)+a(4)+...+a(100)
=2+3*2+3*2^2+...+3*2^(49) + 3 + 3*2 + ...+ 3*2^(49)
=-1+3+3*2+3*2^2+...+3*2^(49) + 3+3*2+...+3*2^(49)
=-1 +2[3+3*2+3*2^2+...+3*2^(49)]
=-1+2*3[1+2+2^2+...+2^(49)]
=-1+2*3[2^(50)-1]
=3*2^(51) - 7