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

 1, n,...,n,1
The average of the entries in this row is 2048. Find n

 

i originally got 2047 but that was wrong

 Dec 18, 2023
 #1
avatar+56 
0

 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.

 Dec 18, 2023
 #2
avatar+14 
+1

Sorry, that's wrong 

 Dec 19, 2023
 #3
avatar+126665 
+1

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

 

cool cool cool

 Dec 19, 2023
edited by CPhill  Dec 19, 2023

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