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The entries in a certain row of Pascal's triangle are
\[1, n, \dots, n, 1.\]
The average of the entries in this row is 2.  Find $n$.

 Oct 6, 2023
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Let the row of Pascal's triangle be [1, n, \binom{n}{2}, \dots, \binom{n}{2}, n, 1.]Then the average of the entries in this row is [\frac{1 + n + \binom{n}{2} + \dots + \binom{n}{2} + n + 1}{n + 1} = \frac{1 + 2(1 + \binom{n}{2}) + 1}{n + 1} = \frac{4 \binom{n}{2} + 3}{n + 1}.]Since the average is 2, we have [\frac{4 \binom{n}{2} + 3}{n + 1} = 2,]so 4(2n)+3=2(n+1). Solving for n, we get n=6.

 Oct 6, 2023

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