Question

If k is odd, then ${}^k{C}_r$ is maximum for r equal to

• A. $\frac{1}{2}(k-1)$
• B. $\frac{1}{2}(k+1)$
• C. $k-1$
• D. $k$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{2}(k-1)$
B $\frac{1}{2}(k+1)$

$\because$ Since $k$ is an odd number.

$\therefore k=2n+1$

Now, ${}^k{C}_r$ becomes ${}^{2n+1}{C}_r$

Now, condition for ${}^{2n+1}{C}_r$ to be max is when $r=n$

but $\therefore {}^n{C}_r={}^n{C}_{2n+1-n}={}^{2n+1}{C}_{n+1}$

Or $r=n$ or $r=n+1$

but $k=2n+1$ , $k=2n+1$

or $k=2r+1$ , $k=2(r-1)+1$

$\Rightarrow$ $r=\frac{k-1}{2}$ , $k=2r-1$

$\Rightarrow$ , $r=\frac{k+1}{2}$

Hence, the answer is $\frac{1}{2}\left(k-1\right),\frac{1}{2}\left(k+1\right).$