若n的平方+3n=1,求n(n+1)(n+2)(n+3)+1的值

若n的平方+3n=1,求n(n+1)(n+2)(n+3)+1的值
若n的平方+3n=1,求n(n+1)(n+2)(n+3)+1的值

若n的平方+3n=1,求n(n+1)(n+2)(n+3)+1的值
因为n2+3n=1
所以n(n+3)=1
所以n(n+1)(n+2)(n+3)+1
=n(n+3)(n+2)(n+1)+1
=(n+2)(n+1)+1
=(n2+2n+n+2)+1
=n2+3n+2+1
=1+2+1
=4

因为n(n+1)(n+2)(n+3)+1
=(n^2+3n)[(n+1)(n+2)]+1
=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n)^2+2(n^2-3n)+1 (1)
把n^2+3n=1代入（1）得
1^2+2*1+1=1+2+1=4
所以n(n+1)(n+2)(n+3)+1的值是4

4