How to simplify the summation ...?

Question

How can I simplify that the
(sum (n choose 2k)*3^n ) / 6^n
(sum from k=0 to n/2)

I need to show that this will always equal .5?

Answer 1

first things first.

(sum (n choose 2k)*3^n ) / 6^n =
3^n * sum (n choose 2k) / 6^n =
sum (n choose 2k) / 2^n [0 <= k <= n/2]

and that in effect is same as asking,

"from the set of n elements, how many subsets can we select with even number of elements"?

so now if we show that # of even-length subsets equals the number of odd-length subsets, we'll have shown that the sum in question equals 2^n/2, and the solution is 0.5

so

let 0 denote an element not belonging to a subset, and let 1 denote an element belonging to the subset. Let's list all subsets of S as follows:

1: 0000...0000
2: 0000...0001
3: 0000...0010
4: 0000...0011
5: 0000...0100
6: 0000...0101

etc. Each string above is n-digits long, and the total number of such strings is 2^n.

Now let's compare two strings in positions 2k-1, 2k. The string in position 2k-1 ends with a 0, string in position 2k ends with a 1. There are no other differences between the two strings. So, if one of them had an even number of 1's, the other must have an odd number of 1's.

This way, we show that every subset of S with even number of elements has a corresponding subset with odd number of elements. Thus,

sum (n choose 2k) = 2^n / 2, and
sum (n choose 2k) / 2^n = 0.5 indeed.

cool problem!

Answer 2

3^n//6^n= (1/2)^n=
(1/2)^2k =
The sum from k= 0 to n/2 is
1+1/2^2+1/2^4 ++++ 1/2^2k=
This is the sum of a geometric progression with r=1/4
S=(1/2^2k*1/4-1)/1/4-1 = (1/2^n*1/4-1)/(1/4-1)
which as n=> infinity tends to 4/3
As far as I understand you your question
As the sum begins with 1 it never can be.5 and will depend on n