Question

一道开根号的题目根号里面是 2004×2005×2006×2007+1

Answer 1

n(n+1)(n+2)(n+3)+1

=(n^2+3n)(n^2+3n+2)+1

=[(n^2+3n+1)-1][(n^2+3n+1)+1]+1

=[(n^2+3n+1)^1-1]+1

=(n^2+3n+1)^2,

√[n(n+1)(n+2)(n+3)+1]=n^2+3n+1,

n=2004时,得到

√[2004*2005*2006*2007+1}

=2004^2+3*2004+1

=4022029.