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
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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](/uploads/image/z/11588860-28-0.jpg?t=1%2A2%2A3%2A4%2B1%3D5%5E2%2C2%2A3%2A4%2A5%2B1%3D11%5E2%2C3%2A4%2A5%2A6%2B1%3D19%5E2%2C...%E5%86%99%E5%87%BA%E4%B8%80%E4%B8%AA%E5%85%B7%E6%9C%89%E6%99%AE%E9%81%8D%E6%80%A7%E7%9A%84%E7%BB%93%E8%AE%BA%2C%E5%B9%B6%E7%BB%99%E5%87%BA%E8%AF%81%E6%98%8E)
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