pascal's triangle

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​.