Question

求积(1+1/1*3)(1+1/2*4)(1+1/3*5)......(1+1/99*101)的整数部分

Answer 1

1+1/1*3=4/1*3=2^2/1*3
1+1/2*4=9/2*4=3^2/2*4
(1+1/1*3)(1+1/2*4)(1+1/3*5)......(1+1/99*101)
=(2^2/1*3)(3^2/2*4)(4^2/3*5)...(100^2/99*101)
=(2^2*3^2*4^2*...*100^2)/(1*2*3^2*4^2*...99^2*100*101)
=2*100/101
=200/101
=1+99/101
整数部分为1

Answer 2

式子1+1/1*3=(2/1)*(2/3),
1+1/2*4=(3/2)*(3/4).
(1+1/1*3)(1+1/2*4)(1+1/3*5)......(1+1/99*101)
=[(2/1)*(2/3)]*[(3/2)*(3/4)]*[(4/3)*(4/5)]……[(100/99)*(100/101)]
=2*【(2/3)*(3/2)】*[(3/4)*(4/3)]*[4/5…………（100、99)]*(100/101)
=200/101
=1.98