If $$^{2n+1}P_{n-1} : ^{2n-1}P_n = 11 : 21$$, then $$n^2 + n + 15$$ is equal to:

$$\frac{^{2n+1}P_{n-1}}{^{2n-1}P_{n}} = \frac{11}{21}$$

$$\frac{\frac{(2n+1)!}{(2n+1-(n-1))!}}{\frac{(2n-1)!}{(2n-1-n)!}} = \frac{11}{21} \implies \frac{(2n+1)!}{(n+2)!} \cdot \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$$

$$\frac{(2n+1)(2n)(2n-1)!}{(n+2)(n+1)(n)(n-1)!} \cdot \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$$

$$\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21} \implies 42(2n+1) = 11(n^2+3n+2)$$

$$84n + 42 = 11n^2 + 33n + 22 \implies 11n^2 - 51n - 20 = 0$$

$$(11n + 4)(n - 5) = 0 \implies n = 5$$

$$n^2 + n + 15 = 5^2 + 5 + 15 = 45$$