Question #251145

Prove that for all integer n3, P(n+1,3)−P(n,3)=3P(n,2).


1
Expert's answer
2021-10-15T09:34:45-0400

We shall prove by mathematical induction.

For n=3,n=3,

P(4,3)P(3,3)=246=18=3(6)=3P(3,2)P(4,3)-P(3,3)=24-6=18=3(6)=3P(3,2)

Thus it is true for n=3n=3

Assume it is true for n=k,k3.n=k,k\geq3. Then

P(k+1,3)P(k,3)=3P(k,2)P(k+1,3)-P(k,3)=3P(k,2)

We shall show that it is true for n=k+1,k3n=k+1,k\geq 3

That is, P(k+2,3)-P(k+1,3)=3P(k+1,2)

P(k+2,3)P(k+1,3)=(k+2)!(k1)!(k+1)!(k2)!=k(k+1)(k+2)k(k1)(k+1)=3k(k+1)P(k+2,3)-P(k+1,3)=\frac{(k+2)!}{(k-1)!}-\frac{(k+1)!}{(k-2)!}=k(k+1)(k+2)-k(k-1)(k+1)\\ =3k(k+1)

For the RHS

3P(k+1,2)=3((k+1)!(k1)!)=3k(k+1)3P(k+1,2)=3\left(\frac{(k+1)!}{(k-1)!}\right)=3k(k+1)

Since LHS=RHS, then it is true for n=k+1,k3n=k+1,k\geq 3

Hence, P(n+1,3)P(n,3)=3P(n,2),n3P(n+1,3)-P(n,3)=3P(n,2), n\geq 3


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