If {}^{2n+1}P_{n-1} : {}^{2n-1}P_n = 11 : 21, | vedclass

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(D) Given the ratio: $\frac{{}^{2n+1}P_{n-1}}{{}^{2n-1}P_n} = \frac{11}{21}$Using the formula ${}^{n}P_r = \frac{n!}{(n-r)!}$,we have:$\frac{(2n+1)!}{

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$45$

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(D) Given the ratio: $\frac{{}^{2n+1}P_{n-1}}{{}^{2n-1}P_n} = \frac{11}{21}$Using the formula ${}^{n}P_r = \frac{n!}{(n-r)!}$,we have:$\frac{(2n+1)!}{(2n+1 - (n-1))!} 	imes \frac{(2n-1 - n)!}{(2n-1)!} = \frac{11}{21}$$\frac{(2n+1)!}{(n+2)!} 	imes \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$Expanding the factorials:$\frac{(2n+1)(2n)(2n-1)!}{(n+2)(n+1)n(n-1)!} 	imes \frac{(n-1)!}{(2n-1)!} = \frac{11}{21}$$\frac{(2n+1)(2n)}{(n+2)(n+1)n} = \frac{11}{21}$$\frac{2(2n+1)}{(n+2)(n+1)} = \frac{11}{21}$$42(2n+1) = 11(n^2 + 3n + 2)$$84n + 42 = 11n^2 + 33n + 22$$11n^2 - 51n - 20 = 0$Solving the quadratic equation $11n^2 - 55n + 4n - 20 = 0$:$11n(n-5) + 4(n-5) = 0$$(11n+4)(n-5) = 0$Since $n$ must be a positive integer,$n = 5$.Calculating $n^2 + n + 15$:$5^2 + 5 + 15 = 25 + 5 + 15 = 45$.

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

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