Question

If ^n–1C_r = (k² – 8) ⁿC_r+1, then the range of 'k' is

(1) \(k \in (2\sqrt{2},3]\)

(2) \(k\in (2\sqrt{2},3)\)

(3) \(k\in [2,3]\)

(4) \(k\in (2\sqrt{2},8)\)

Answer 1

(1) \(k \in (2\sqrt{2},3]\)

\(^{n-1}C_r =(k^2 -8)\frac {n}{r +1}.^{n-1}C_r\)

⇒ \(k^2 -8 =\frac {r+1}{n}\)

here r \(\in\) [0, n – 1] 

⇒ r + 1 \(\in\) [1, n]