Question

已知数列a(n)的前n项和为s(n),a(n)等于(2n-1)(4n+2)分之1,求s(n)?

Answer 1

an=1/(2n-1)(4n+2)=1/2*1/(2n-1)(2n+1)=1/2*1/2*[1/(2n-1)-1/(2n+1)]

Sn=1/4(1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1))

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

=1/4*2n/(2n+1)

=n/[2(2n+1)]