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At the very last step, highlighted in red, it states that if this, then that sort of, and I'm not sure how those comparisons explain the issue of proving Pascal's Triangle increasing until the middle, then decreases.

Is it saying (for the first one) that if the value of one entry is less than the next, then the entry comes before the next one, and vice-versa for greater than? If this is the case, how does that make sense for equals? If the value of one entry is greater than the next, the two occur in the same spot? How can two occur in the same spot? (I know in odd rows the middle is duplicated, but they're still difference positions within the row.)

Could anyone offer an explanation?

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up vote 3 down vote accepted

What the author is doing is this:

Comparing successive values in a row of Pascal's triangle, the terms are increasing so long as we have $1 < \frac{n-k}{k+1}$, which is the same thing as $k < \frac{n-1}{2}$.

This means that until $k$ gets to be as big as $\frac{n-1}{2}$, the entries are increasing, and if this fraction is equal to 1, then we get 2 identical terms in the middle of the row. When we have $k$ bigger than $\frac{n-1}{2}$ (i.e. after the middle entry/entries of the row), then the entries are decreasing.

Does this make the implications clearer?

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So when you say 'until k gets to be as big as $\frac{n-1}{2}$', what exactly is that saying? Until the value of some item is as big as the middle (because if n is the row, $\frac{n-1}{2}$ is the middle entry of the row, as I'm sure you know)? I guess I'm just confused whether or not $k$ is referring to a value of an entry, or its position in the row. $\binom{n}{k}$ is compared against $\binom{n}{k+1}$, does that compare one entry against the next? And if that's the case, does $k < \frac{n-1}{2}$ represent one entry being elss than the next? – Doug Smith Nov 7 '12 at 1:40
$k$ is the number of the entry in that particular row in the triangle. $k=0$ is the first entry in the row, $k=1$ is the second entry in the row etc. So, when $k$ is less than $(n-1)/2$ we are looking at the entries in the row before the middle. – Old John Nov 7 '12 at 1:44
So when $k = \frac{n-1}{2}$, k is at the middle? So if $k = \frac{n-1}{2}$, does that just mean when k is at the middle, it's equal to the middle? Is that not a redundant statement? – Doug Smith Nov 7 '12 at 2:11
@DougSmith The $k$th entry of the row is to the left of the $k+1$th entry, for each $k$. If $n$ is even, $(n-1)/2$ is strictly just before the last index of those indices of the first half of the row. If $n$ is odd, $(n-1)/2$ is the index of the entry just before the middle entry. Test out some actual numbers if you need to in order to follow along. Either way, if the successive terms are increasing on the LHS of the row and decreasing on the RHS, then of course the middle entry (or the middle two entries if $n$ is odd) will be maximal among the row's entry. – anon Nov 7 '12 at 2:22
Ah, fair enough, thanks so much. – Doug Smith Nov 7 '12 at 2:23

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