（探究题）观察思考：1×2×3×4＋1＝25＝5²2×3×4×5＋1＝121＝11²3×4×5×6＋

这是第一题
首先 设第一个数为 n
推的结论 ：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
即n(n+1)(n+2)(n+3)+1 =(n^2+3n+1)^2
证明 （写的不太好 将就看吧）则有
n(n+1)(n+2)(n+3)+1=[n(n+3)+1]^2
n(n+3)(n+1)(n+2)+1=n^2(n+3)^2+2n^2+6n+1
n(n+3)(n^2+3n+2)+1=n^2(n+3)^2+2n^2+6n+1
(n^2+3n)(n^2+3n+2)+1=n^2(n^2+6n+9)+2n^2+6n+1
n^4+3n^3+2n^2+3n^3+9n^2+6n+1=n^4+6n^3+9n^2+2n^2+6n+1
6n+1=6n+1
为恒等式 所以成立