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

\frac{1}{2} (k - 1)

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\frac{1}{2} (k + 1)

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k - 1

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k

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Solution

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

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Similar questions

\infty \sum k = 1 \frac{6^{k}}{(3^{2 k + 1} + 2^{2 k + 1}) - (3^{k} 2^{k + 1} + 2^{k} 3^{k + 1})}

(k, 2 k), (2 k, 3 k), (3, 1)

k

A = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 2 k - 1 & 2 \sqrt{k} & 2 \sqrt{k} 2 \sqrt{k} & 1 & - 2 k - 2 \sqrt{k} & 2 k & - 1 \end{matrix} ⎤ ⎥ ⎥ ⎦

B = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 0 & 2 k - 1 & \sqrt{k} 1 - 2 k & 0 & 2 \sqrt{k} - \sqrt{k} & - 2 \sqrt{k} & 0 \end{matrix} ⎤ ⎥ ⎥ ⎦

(A d j (A)) + d e t (A d j (B)) = 10^{o}

[k]

A B ⊥ B C

k

A (k, 1, - 1), B (2 k, 0, 2), C (2 + 2 k, k, 1)

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