If ^2 n C_3: ^n C_3=10: 1, then the ratio [(n^2+3 n):(n^2-3 n+4)] is :

Question

If ${ }^{2 n} C_{3}:{ }^{n} C_{3}=10: 1$, then the ratio $$\left(n^{2}+3 n\right):\left(n^{2}-3 n+4\right)$$ is :

Accepted Answer

$$
\begin{aligned}
& \text {We have, }{ }^{2 n} C_3:{ }^n C_3=10: 1 \\
& \Rightarrow \frac{{ }^{2 n} C_3}{{ }^n C_3}=\frac{10}{1} \\
& \Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=\frac{10}{1} \\
& \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{(n)(n-1)(n-2)}=\frac{10}{1} \\
& \Rightarrow 4 n^2-6 n+2=5\left(n^2-3 n+2\right) \\
& \Rightarrow n^2-9 n+8=0 \\
& \Rightarrow n^2-8 n-n+8=0 \\
& \Rightarrow n(n-8)-1(n-8)=0 \\
& \Rightarrow (n-8)(n-1)=0 \\
& \Rightarrow n=8(n=1 \text { not valid })
\end{aligned}
$$

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

Solutions · 1