JEE Main 2023 — Mathematics Algebra

We have been given that $C32n:C3n=10$C32n:C3n=10

We know that $Crn=\frac{n!}{r!(n-r)!}$Crn=r!(n−r)!n!​

$$\Rightarrow \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}=10$$⇒3!(n−3)!n!​3!(2n−3)!(2n)!​​=10

$$\Rightarrow \frac{(2n)!}{3!(2n-3)!}\times \frac{3!(n-3)!}{n!}=10$$⇒3!(2n−3)!(2n)!​×n!3!(n−3)!​=10

$$\Rightarrow \frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!}\times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!}=10$$⇒(2n−3)!(2n)(2n−1)(2n−2)(2n−3)!​×n(n−1)(n−2)(n−3)!(n−3)!​=10

$$\Rightarrow \frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)}=10$$⇒n(n−1)(n−2)(2n)(2n−1)(2n−2)​=10

$$\Rightarrow \frac{(4)(2n-1)}{(n-2)}=10$$⇒(n−2)(4)(2n−1)​=10

$$\Rightarrow 8n-4=10n-20$$⇒8n−4=10n−20

$$\Rightarrow n=8$$⇒n=8

Hence, the value of $\frac{{n}^{2}+3n}{{n}^{2}-3n+4}$n2−3n+4n2+3n​ is $=\frac{{8}^{2}+3(8)}{{8}^{2}-3(8)+4}=\frac{88}{44}=2$=82−3(8)+482+3(8)​=4488​=2

Therefore, the value of $\frac{{n}^{2}+3n}{{n}^{2}-3n+4}$n2−3n+4n2+3n​ is $2:1$2:1.