n是奇数则有(n-1)/2个偶数项q=2^2=4,首项2^2=4所以和=4*[4^(n-1)/2-1]/(4-1)=(4/3)*[2^(n-1)-1]有(n+1)/2个奇数项a1=1,an=2n-1所以和=(1+2n-1)*[(n+1)/2]/2=n(n+1)/2n是偶数则有n/2个偶数项q=2^2=4,首项2^2=4所以和=...