1234+1=5^2,2345+1=11^2,345*6+1=19^2,...写出一个具有普遍性的结论,并给出证明

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1*2*3*4+1=5^2,2*3*4*5+1=11^2,3*4*5*6+1=19^2,...写出一个具有普遍性的结论,并给出证明1*2*3*4+1=5^2,2*3*4*5+1=11^2,3*4*5*

1*2*3*4+1=5^2,2*3*4*5+1=11^2,3*4*5*6+1=19^2,...写出一个具有普遍性的结论,并给出证明
1*2*3*4+1=5^2,2*3*4*5+1=11^2,3*4*5*6+1=19^2,...写出一个具有普遍性的结论,并给出证明

1*2*3*4+1=5^2,2*3*4*5+1=11^2,3*4*5*6+1=19^2,...写出一个具有普遍性的结论,并给出证明
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+1)-1][(n^2+3n+1)+1]+1
=(n^2+3n+1)^2-1+1
=(n^2+3n+1)^2.

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

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

你好！

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

1+3-5*4^2-3= 1+2+3-4-5+1+1+2+3= 1 2 3 4 5=1 1()2()3()4()5=1 1+1-2+3-4+5=? 1/2+2/3+3/4+4/5=? 1/2+1/(2+3)+1/(2+3+4)+.+1/(2+4+5+...)200)=? 1+2=3 4+5=? 1/2+1/3+2/3+1/4+2/4+3/4+1/5+2/5+3/5+4/5+...+1/100+2/100+...+99/100=? 1/2+(1/3+2/3)+(1/4+2/4+3/4)+(1/5+2/5+3/5+4/5)+...+(1/60+2/60+.+59/60)=? 1+2+3+4+5+.+99=? 1+2+3+4+5+.+99=? 1+2+3+4+5+.10000= 2+3+4+5+.+n-1=?, 1+2-3+4-5+6= 1+2+3+4+5-------------+1000=? 1+2+3+4+5+.+50=? 是不是5*4*3*2*1=?