一道规律题

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

证明应该不难的
n(n+1)(n+2)(n+3)+1
=(n^2+3n)*(n^2+3n+2)+1
=(n^2+3n+2)^2-2(n^2+3n+2)+1
=(n^2+3n+2-1)^2
=[(n+1)(n+2)-1]^2
=(n^2+3n+1)^2

根号(0*1*2*3+1)=1=0*3+1
根号(1*2*3*4+1)=5=1*4+1
根号(2*3*4*5+1)=11=2*5+1
......
根号(6*7*8*9+1)=55=6*9+1
......
用含有n的式子表示你发现的规律,并证明你发现的规律``

根号[n(n+1)(n+2)(n+3)+1]=n(n+3)+1

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

所以根号[n(n+1)(n+2)(n+3)+1]=根[n(n+3)+1]^2=n(n+3)+1