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  • If <img alt="\mathrm{{ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_n=11: 21, then \: n^2+n+15}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%7B%20%7D%5E%7B2%20n&plus;

If \mathrm{{ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_n=11: 21, then \: n^2+n+15} is equal to :
 

Option: 1

45


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{ { }^{2 n+1} P_{n-1}:{ }^{2 n-1} P_n=11: 21 }
\mathrm{ \frac{(2 n+1) !}{(n+2) !} \times \frac{(n-1) !}{(2 n-1) !}=\frac{11}{21} }
\mathrm{ \Rightarrow \frac{(2 n+1)(2 n)}{(n+2)(n+1) n}=\frac{11}{21} }
\mathrm{ \Rightarrow \frac{2(2 n+1)}{(n+2)(n+1)}=\frac{11}{21} }
\mathrm{ \Rightarrow 42(2 n+1)=11\left(n^2+3 n+2\right) }
\mathrm{ \Rightarrow 84 n+42=11 n^2+33 n+22 }
\mathrm{ \Rightarrow 11 n^2-51 \mathrm{n}-20=0 }
\mathrm{ \Rightarrow \mathrm{n}=5 }
\mathrm{ \mathrm{n}^2+\mathrm{n}+15=25+5+15=45 }
 

Posted by

Ritika Kankaria

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