Since , there will be terms in the expansion, which means that there will be middle terms in the expansion, the and the :
Is there any term in the expansion of that will be independent of ?
The general term in the expansion is
Thus, for the term that is independent of , we have
Thus, the term free of is the term given by
Evaluate the sum .
We have already evaluated this sum in the chapter on & . That approach was as follows: this sum basically counts the number of all sub-groups of a set of size ; this can also be counted by focusing on each element of the set, which has two corresponding choices – you either include it into your sub-group or you don’t, which means that the total number of ways to form sub-groups is . The sum of the binomial coefficients therefore equals .
Here, we evaluate the same sum using a binomial approach. Consider the following expansion:
If we put , we simply obtain
Thus, the same result is obtainable from both a combinatorial and a binomial approach.
We can also derive another useful result by putting in the above relation, so that we obtain
This states the sum of the even-numbered coefficients is equal to the sum of the odd-numbered coefficients. Can you prove this using a combinatorial approach?