Question

设平面内n条直线把平面分成最多块数为f(n),求证:f(n)=(n²+n+2)(n∈N^*).

Answer 1

证明：(1)当n=1时,f(1)=(1+1+2)=2,命题成立.

(2)假设n=k时等式成立,即f(k)=(k²+k+2).

当n=k+1时,第k+1条直线与其余k条直线都相交且任意三线不共点,则比k条线时增加(k+1)块.

∴f(k+1)=f(k)+k+1=(k²+k+2)+k+1

=(k²+2k+1+k+3)

=［(k+1)²+(k+1)+2］.

∴当n=k+1时等式也成立.

由(1)(2)可知f(n)=(n²+n+2)(n∈N^*).