n(n+1)(n+2)(n+3)+1
=[n(n+3)][(n+1)(n+2)]+1
=[n²+3n][n²+3n+2]+1
=(n²+3n)²+2(n²+3n)+1
=(n²+3n+1)²
=[n(n+3)][(n+1)(n+2)]+1
=[n²+3n][n²+3n+2]+1
=(n²+3n)²+2(n²+3n)+1
=(n²+3n+1)²
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第1个回答 2008-08-31
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+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
第2个回答 2008-08-31
n(n+1)(n+2)(n+3)+1
=[n(n+3)][(n+2)(n+3)]+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+3)][(n+2)(n+3)]+1
=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2
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