Question

若n的平方+3n=1,求n(n+1)(n+2)(n+3)+1的值

Answer 1

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