If $k\geq 0$ and $k\leq\frac{n-1}{2}$, then prove that $$\binom{n}{k}\leq \binom{n}{k+1}$$.
Then prove there is equality only if $k=\frac{n-1}{2}$.
Induction with Binomial Coefficients
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combinatorics
induction
binomial-coefficients
2 Answers
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$0\leq k\leq\frac{n-1}{2} \implies 2k \leq n-1 \implies k+1 \leq n-k \implies \frac{1}{ n-k} \leq \frac{1}{1+k}$
$$\frac{1}{ n-k} \leq \frac{1}{1+k} \implies \frac{n!}{ (n-k)k!(n-k-1)!} \leq \frac{n!}{(1+k)k!(n-k-1)!} \\ \implies \binom{n}{k}\leq \binom{n}{k+1}$$
Equality can be verified very easily.
Just put $k=\frac{n-1}{2}$ in the binomial coefficient to see that they are equal.
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0Ok, so the first line is the first part, and the rest is to prove equality? – 2017-02-17
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0can you please explain i did not understand – 2017-02-17
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0Did you prove the equality in the above answer? – 2017-02-17
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$\begin{array}\\ r_k &=\dfrac{ \binom{n}{k+1}}{\binom{n}{k}}\\ &=\dfrac{ \frac{n!}{(k+1)!(n-k-1)!}}{\frac{n!}{k!(n-k)!}}\\ &=\dfrac{ k!(n-k)!}{(k+1)!(n-k-1)!}\\ &=\dfrac{n-k}{k+1}\\ \end{array} $
Therefore $r_k \le 1 \iff n-k \ge k+1 \iff k \le \frac{n-1}{2} $ with equality if and only of $k = \frac{n-1}{2} $.