If ^2nC_3:^nC_3=10:1, then the ratio (n^2+3n):(n^2-3n+4) is :

Question

If ${}^{2n}{C}_{3}:{}^{n}{C}_{3}=10:1$, then the ratio $({n}^{2}+3n):({n}^{2}-3n+4)$ is :

Accepted Answer

$\begin{gathered} {}^{2n}{C}_{3}:{}^{n}{C}_{3}=10:1 \ \Rightarrow \frac{{}^{2n}{C}_{3}}{{}^{n}{C}_{3}}=\frac{10}{1} \ \Rightarrow \frac{(2n)!}{3!(2n-3)!}	imes \frac{3!(n-3)!}{n!}=\frac{10}{1} \ \Rightarrow \frac{(2n)(2n-1)(2n-2)}{(n)(n-1)(n-2)}=\frac{10}{1} \ \Rightarrow 4{n}^{2}-6n+2=5({n}^{2}-3n+2) \ \Rightarrow {n}^{2}-9n+8=0 \ \Rightarrow {n}^{2}-8n-n+8=0 \ \Rightarrow n(n-8)-1(n-8)=0 \ \Rightarrow (n-8)(n-1)=0 \ \Rightarrow n=8(n=1) \end{gathered}$

$	herefore \frac{{n}^{2}+3n}{{n}^{2}-3n+4}=\frac{88}{44}=\frac{2}{1}=2:1$

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