Find maximum and minimum value of: $\binom{16}{k}+\binom{16}{k-1},k\in N$
Now, $\binom{16}{k}+\binom{16}{k-1}$ is same as $\binom{17}{k}$
I think that minimum value would be for $k=17$, which is $1$. But what would be the maximum value?
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4As $k$ gets closer to $\frac{17}2$, $\binom{17}k$ gets bigger – J. W. Tanner Nov 03 '20 at 23:44
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1"as $k$ gets closer to $\frac{17}{2}$" of course is in specific reference to the $17$ in $\binom{17}{k}$. More generally $\binom{n}{k}$ is maximized for $k$ closest to $\frac{n}{2}$. – JMoravitz Nov 03 '20 at 23:46
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1Look up Pascal's triangle and find the maximum value of ${3 \choose k}$, ${4 \choose k}, {5 \choose k}$ and so on. – Toby Mak Nov 03 '20 at 23:48
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Cf. this question – J. W. Tanner Nov 03 '20 at 23:49
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2If you are allowed $k=17$ so $\binom{16}{17}+\binom{16}{17-1}=1$, then $k=18$ and $\binom{16}{18}+\binom{16}{18-1}=0$ is an even smaller result – Henry Nov 03 '20 at 23:54