Question

等差数列前n项和问题求和：1+1/1+2+1/1+2+3+…+1/1+2+3+…+n

Answer 1

1+1/1+2+1/1+2+3+…+1/1+2+3+…+n

=1+1/3+1/6+.+1/(1+n)n/2

=1+1/3+1/6+.2/(1+n)n

=2[1/2+1/6+1/12+.+1/n(n+1)]

=2[1/1*2+1/2*3+1/3*4+.+1/n(n+1)]

=2[1-1/2+1/2-1/3+1/3-1/4+...1/n-1/(n+1)]

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

=2*n/(n+1)

=2n/(n+1)