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 $n$.

blackpanther Feb 13, 2024

#1**+2 **

The entries in row n of Pascal's triangle are 1, n,..., n, 1. Additionally, __the sum of row n's elements is 2^n__.

Additionally, __row n has n + 1 terms__, so we know from the problem that \({2^n\over{n+1}}=2\).

Cross multiplying, you get \(2^n=2n+2\). The question here is trivial by bash => you will find that __ n = 3__ works. (If you are a quick guess + checker, you might've found that right of the bat quickly, as row 3 is 1, 3, 3, 1 which has an average entry of 2.)

proyaop Feb 13, 2024

#1**+2 **

Best Answer

The entries in row n of Pascal's triangle are 1, n,..., n, 1. Additionally, __the sum of row n's elements is 2^n__.

Additionally, __row n has n + 1 terms__, so we know from the problem that \({2^n\over{n+1}}=2\).

Cross multiplying, you get \(2^n=2n+2\). The question here is trivial by bash => you will find that __ n = 3__ works. (If you are a quick guess + checker, you might've found that right of the bat quickly, as row 3 is 1, 3, 3, 1 which has an average entry of 2.)

proyaop Feb 13, 2024