QJEEM23Mace0d: JEE Main 2023 Mathematics - Class 11 JEEM23Mace0d Solution: If , then is equal to :

Question

Verified Solution • 2025–26 Session moderate

If

{ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_{n}=11: 21

,

then

n^{2}+n+15

is equal to :

Accepted Answer

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

\frac{(2 n+1) 2 n}{(n+2)(n+1) n}=\frac{11}{21}

84 n+42=11\left(n^{2}+3 n+2\right)

11 n^{2}-51 n-20=0

(n-5)(11 n+4)=0

n=5, \frac{-4}{11}

(Rejected

)

n^{2}+n+15=45

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