Question

if 2n+1pn-1:2n-1pn =3:5 find n

Answer 1

telecommunications paraprofessional specialists fan

Answer 2

$\huge\sf{\underline{\underline{Correct\:\:Question:}}}$

$\rm \: If \: {}^{2n + 1} \: P_n - 1 \: \colon \: {}^{2n - 1} \: P_n = 3 \: \colon \: 5$

$\huge\sf{\underline{\underline{Answer:}}}$

We have,

$\sf \qquad \: {}^{2n + 1} \: P_n - 1 \: \colon \: {}^{2n - 1} \: P_n = 3 \: \colon \: 5 \\ \\ \rightarrow \sf \: \frac{ {}^{2n + 1} \: P_n - 1 }{ {}^{2n - 1} \:P_n } = \frac{3}{5} \\ \\ \rightarrow \sf \frac{(2n + 1)!}{(n + 2)!} \times \frac{(n - 1)!}{(2n - 1)!} = \frac{3}{5} \\ \\ \rightarrow \sf \frac{(2n + 1)(2n)(2n - 1)!}{(n + 2)(n + 1)n(n - 1)!} \times \frac{(n - 1)!}{(2n - 1)!} = \frac{3}{5} \\ \\ \rightarrow \sf \frac{2(2n + 1)}{(n + 2)(n + 1)} = \frac{3}{5} \\ \\ \rightarrow \sf \: 10(2n + 1) = 3(n + 2)(n + 1) \\ \\ \rightarrow \sf3 {n}^{2} + 9n + 6 = 20n + 10 \\ \\ \rightarrow \sf \: 3 {n}^{2} - 11n - 4 = 0 \\ \\ \rightarrow \sf \: (n - 4)(3n + 1) = 0 \\ \\ \rightarrow \sf \: n = 4$