Question

100分,答求1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+.+100)的值

Answer 1

1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+.+100)

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

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

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

=2-2/(n+1)

原式为n=100时的数值

即原式=2-2/101=200/101