Question

求和：Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)

Answer 1

Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)

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

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

=n+[(1-1/3)/2+(1/3-1/5)/2+...+(1/(2n-1)-1/(2n+1)/2]

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

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

=n+n/(2n+1)