1958 AHSME Problems/Problem 49
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Problem
In the expansion of there are
dissimilar terms. The number of dissimilar terms in the expansion of
is:
Solution
Expand the binomial with the binomial theorem. We have:
So for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:
Solution 2 (Stars and Bars)
Each term in the expansion of will have the form
, where
and
. We just need to find the number of triplets of nonnegative integers
such that
. Thus, our answer is
=66.
See also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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