√2004*2005*2006*2007+1数字全在根号下
问题描述：
√2004*2005*2006*2007+1
数字全在根号下
答
n(n+1)(n+2)(n+3)+1=(n^2+3n)(n^2+3n+2)+1=(n^2+3n+1)(n^2+3n+1)
将n=2004代入
答
√(2004×2005×2006×2007+1)
为方便起见,设n=2004,所以原式成为：
√(2004×2005×2006×2007+1)
=√[n(n+1)(n+2)(n+3)+1]
=√[n(n+3)×(n+1)(n+2)+1]
=√[(n^2+3n)(n^2+3n+2)+1]
=√[(n^2+3n)^2+2(n^2+3n)+1]
=√[(n^2+3n)+1]^2
=√(n^2+3n+1)^2
=n^2+3n+1
=n(n+3)+1
=2004×(2004+3)+1
=2004×2007+1
=4022028+1
=4022029