Question

If n-1Cr = (k2-3) nCr +1 , Prove that k lies between root of 3 and 2 i.e (root of 3 , 2)

Answer 1

your question is ---> $^{n-1}C_r=(k^2-3)^nC_{r+1}$

use formula, $^AC_B=\frac{A!}{B!(A-B)!}$

(n - 1)!/{r! × (n - 1 - r)!} = (k² - 3) × n!/(r + 1)!(n - r - 1)!

or, (n - 1)!/r! = (k² - 3) × n!/(r + 1)!

or, (n - 1)!/r! = (k² - 3) × n(n - 1)!/(r + 1)r!

or, 1/1 = (k² - 3) × n/(r + 1)

or, (r + 1)/n = (k² - 3)

we know, r and n are integers so, (r + 1)/n $\in$ (0, 1]

so, 0 < (r + 1)/n ≤ 1

or, 0 < k² - 3 ≤ 1

or, 3 < k² ≤ 4

or, √3 < k ≤ 2 , -2 ≤ k < -√3

hence, k $\in$ (√3, 2]

Answer 2

Answer:

Step-by-step explanation:

$(n-1){C_r}=(k^2-3).n{C_r+1}$

$\frac{(n-1)!}{r!(n-1-r)!}=(k^2-3)\frac{n!}{(r+1)!(n-r-1)!}$

$\frac{(n-1)!}{r!(n-1-r)!}=(k^2-3)\frac{n.(n-1)!}{r!(r+1)(n-r-1)!}$

$\frac{r+1}{n}=(k^2-3)$

$We\:have$

$r\leq\:(n-1)$

$r+1\leq\:n$

$\frac{r+1}{n}\leq1$

$k^2-3=\frac{r+1}{n}\leq1$

$k^2-3\leq1$

$k^2\leq4$

$-2\leq\:k\leq2........(1)$

Also

$0< k^2 -3$

$=> 3 < k^2\\\\\sqrt{3}<k...........(2)$

From (1) &(2)

$\sqrt{3}<k\leq\:2$

$k\:belongs\:to\:(\sqrt(3), 2]$