P(n,r) is another notation for nPr.
Yes, this is true.
The proof is simply an algebraic expansion, factoring, and verification of an identity.
Expanding, we get n!/ (n-2)! + (n+1)! / (n-1)!
Get a common denominator, (n-1)!; then the numerator becomes
n!(n-1)! + (n+1)! which becomes n!n + (n+1)n! - n!
and then this numerator becomes n!(n + n+1 + 1 - 1) or 2n n!
Finally, then this sum is 2n n! /(n-1)! = 2nn = 2n2