Question:
If $^{n-1}C_r =(k^2 -3) \, ^nC_{r+1}$then k belongs to
If $^{n-1}C_r =(k^2 -3) \, ^nC_{r+1}$then k belongs to
Updated On: Jun 14, 2022
- $(-\infty,-2]$
- $[2,-\infty,)$
- $[-\sqrt 3,\sqrt 3]$
- (\sqrt{3, 2}]
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The Correct Option is D
Solution and Explanation
Given, $^{n-1}C_r=(k^2-3) \, ^nC{r+1}$
= $ \, \, \, \, \, \, \, \, \, \, \, ^{n-1}C_r=(k^2-3)\frac{n}{r+1} \, ^{n-1}C_r$
= $ \, \, \, \, \, \, \, \, \, \, \, k^2-3=\frac{r+1}{n}$
$\hspace20mm [since, n \ge r \Rightarrow \, \, \frac{r+1}{n} \le 1 \, and \, n,r >0]$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, 0< k^2-3\le 1$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, 3 < k^2 \le 4 \, \, \Rightarrow \, \, \, k \in [-2,-\sqrt 3) \cup (\sqrt 3,2)]$
= $ \, \, \, \, \, \, \, \, \, \, \, ^{n-1}C_r=(k^2-3)\frac{n}{r+1} \, ^{n-1}C_r$
= $ \, \, \, \, \, \, \, \, \, \, \, k^2-3=\frac{r+1}{n}$
$\hspace20mm [since, n \ge r \Rightarrow \, \, \frac{r+1}{n} \le 1 \, and \, n,r >0]$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, 0< k^2-3\le 1$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, 3 < k^2 \le 4 \, \, \Rightarrow \, \, \, k \in [-2,-\sqrt 3) \cup (\sqrt 3,2)]$
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.