Question

已知数列{an}的通项公式是an{2n-1(n为奇数)2^n(n为偶数),求数列an的前n项和Sn

Answer 1

n为奇数时 Sn=2*1-1+2*3-1+.+2*n-1+2^2+2^4+.+2^(n-1)

={2*(1+n)*[(n+1)/2]}/2+[2^2(2^(n-1)/2-1)]/(2-1)

=n(n+1)/2+2^((n+3)/2)-4

n为偶数时 Sn=2*1-1+2*3-1+.+2*(n-1)-1+2^2+2^4+.+2^n

=2*(1+n-1)*(n/2)/2+[2^2*(2^(n/2)-1)]/(2-1)

=n(n-1)/2+2^((n+4)/2)-4