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