Question

（2×5+2)(4×7+2)×……×(2004×2007+2)÷(1×4+2)(3×6+2)×……×(2003×2006

Answer 1

构造：an=[2n*(2n+3)+2]/[(2n-1)*(2n+2)+2]=2*(n+1)(2n+1)/[2*n*(2n+1)]=(n+1)/n

（2×5+2)(4×7+2)×……×(2004×2007+2)÷(1×4+2)(3×6+2)×……×(2003×2006+2)

=[(2×5+2)/(1*4+2)]*[(4*7+2)/(3*6+2)]*.*[(2004×2007+2)/(2003×2006+2)]

=(2/1)*(3/2)*..*(1003/1002)

=1003/1

=1003