Question

an=2n,n为奇数,2^n,n为偶数,求此数列的前n项和Sn

Answer 1

n为奇数项,an-a(n-2)=2n-2(n-2)=4,奇数项为首项为2,公差为4的等差数列.
n为偶数项,an/a(n-2)=2^n/2^(n-2)=4,偶数项为首项为4,公比为4的等比数列.
n为偶数时,Sn为n/2项奇数的和 + 前n/2偶数项的和.
Sn=(n/2)a1+(n/2)(n/2-1)*4/2+4*[4^(n/2)-1]/(4-1)
=n^2/2+(4/3)(2^n-1)
n为奇数时,Sn为前（n+1)/2项奇数项的和+(n-1)/2项偶数项的和.
Sn=(n+1)/2a1+[(n+1)/2][(n+1)/2-1]*4/2+4*{4^[(n-1)/2]-1}/(4-1)
=(n+1)^2/2+(2/3)(2^n-2)