Question

A=48*(1/(3^2-4)+1/(4^2-4+~~~1/(100^2-4)的最接近的整数是多少

Answer 1

A=48*(1/(3^2-4)+1/(4^2-4+~~~1/(100^2-4)

=48*[1/(1*5)+1/(2*6)+...+1/(98*102)]

=12*[(1/1-1/5)+(1/2-1/6)+...+(1/98-1/102)]

=12*[1+1/2+1/3+1/4-1/99-1/100-1/101-1/102]

=12+6+4+3-12/99-12/100-12/101-12/102

=25-(12/99+12/100+12/101+12/102).

∵0