If ^2n+1P_n-1:^2n-1P_n=11:21, then n^2+n+15 is equal to : Answer: 45

Question

If ${}^{2n+1}{P}_{n-1}:{}^{2n-1}{P}_{n}=11:21$,

then ${n}^{2}+n+15$ is equal to :

Answer: 45

Accepted Answer

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

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

$84n+42=11({n}^{2}+3n+2)$

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

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

$n=5,\frac{-4}{11}$ (Rejected $)$

${n}^{2}+n+15=45$

Solutions · 1