数列｛an｝的通项公式为an=2n+1,n为奇数,2^n,n为偶数,求此数列的前n项和Sn

Question

数列｛an｝的通项公式为an=2n+1,n为奇数,2^n,n为偶数,求此数列的前n项和Sn

Answer 1

1)当n为奇数2k+1时,则k项偶数和为2^2+2^4+..+2^2k=4(4^k-1)/3则k+1项奇数和为3+5+..+(2k+1)=k(3+2k+1)/2=k(k+2)故Sn=4(4^k-1)/3+k(k+2)2)当n为偶数2k时,n-1=2(k-1)+1则Sn=S(n-1)+a2k=4[4^(k-1)-1]/3]+(k-1)(k+1)+4^k...