If 2 n+1 P n-1: 2 n-1 P n=11: 21, then n2+n+15 is equal to :

Question

If 2 n+1 P n-1: 2 n-1 P n=11: 21, then n2+n+15 is equal to : (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

((2 n +1) !( n -1) !/( n +2) !(2 n -1) !)=(11/21) ⇒ ((2 n +1)(2 n )/( n +2)( n +1) n )=(11/21) ⇒ (2 n +1/( n +1)( n +2))=(11/42) ⇒ n =5 ⇒ n 2+ n +15 =25+5+15=45

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

$\frac{( 2 n + 1 )! ( n - 1 )!}{( n + 2 )! ( 2 n - 1 )!} = \frac{11}{21}$
$\Rightarrow \frac{( 2 n + 1 ) ( 2 n )}{( n + 2 ) ( n + 1 ) n} = \frac{11}{21}$
$\Rightarrow \frac{2 n + 1}{( n + 1 ) ( n + 2 )} = \frac{11}{42}$
$\Rightarrow n = 5$
$\Rightarrow n^{2} + n + 15$
$= 25 + 5 + 15 = 45$

Q. If 2n+1Pn−1​:2n−1Pn​=11:21, then n2+n+15 is equal to :

(n+2)!(2n−1)!(2n+1)!(n−1)!​=2111​ ⇒(n+2)(n+1)n(2n+1)(2n)​=2111​ ⇒(n+1)(n+2)2n+1​=4211​ ⇒n=5 ⇒n2+n+15 =25+5+15=45

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

$\frac{( 2 n + 1 )! ( n - 1 )!}{( n + 2 )! ( 2 n - 1 )!} = \frac{11}{21}$
$\Rightarrow \frac{( 2 n + 1 ) ( 2 n )}{( n + 2 ) ( n + 1 ) n} = \frac{11}{21}$
$\Rightarrow \frac{2 n + 1}{( n + 1 ) ( n + 2 )} = \frac{11}{42}$
$\Rightarrow n = 5$
$\Rightarrow n^{2} + n + 15$
$= 25 + 5 + 15 = 45$