The entries in a certain row of Pascal's triangle are

 Let there be n + 1 entries in the row (including the 1s at the beginning and end). Since the average is 2048, the sum of all entries must be 2048 * (n + 1).

The entries in the given row are consecutive multiples of n, except for the 1s at the ends. Therefore, the sum can be expressed as 1 + \binom{n + 1}{1} + \binom{n + 1}{2} + ... + \binom{n + 1}{n + 1}.

This is an arithmetic series with n + 1 terms and a common difference of n. The sum of an arithmetic series is given by the formula: Sn = n/2 * (a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term.

In this case, a1 = 1 and an = \binom{n + 1}{n + 1}. Plugging these into the formula, we get:

Sn = (n + 1) / 2 * (1 + n^2)

Equating the sums: Now, equate the two expressions for the sum:

2048 * (n + 1) = (n + 1) / 2 * n^2

Cancel out the common factor (n + 1) on both sides:

2048 = (1 + n^2) / 2

Multiply both sides by 2:

4096 = 1 + n^2

Subtract 1 from both sides:

4095 = n^2

Take the square root of both sides:

63 = n

Therefore, n = 63 is the value that makes the average of the entries in the row equal to 2048.

Sorry, that's wrong 

The sum of the entries  in any  row =    2^n      where the number of entries = n + 1   and n = 0,1,2,3,4.......

Therefore

2^n  /  (n + 1)  =   2048

2^n  =  2048 (n + 1)               take the log of both sides

log (2^n)  = log (2048 ( n+ 1) )    using base 2 log we can  write

log₂ (2^n)  = log₂ (2048 (n + 1) )       

n log ₂ 2  = 11 + log₂ (n + 1)         { log ₂ 2   = 1 }

n  = 11 + log₂ (n + 1)

(n - 11)  = log₂ (n + 1)     in exponential form we  have

2^(n -11) = n + 1

Note that this will be true when  n = 15