Problem:
In the expansion of
(1+x+x2+⋯+x27)(1+x+x2+⋯+x14)2
what is the coefficient of x28?
Answer Choices:
A. 195
B. 196
C. 224
D. 378
E. 405
Solution:
Each term in the expansion has the form xa+b+c, where 0≤a≤27,0≤b≤14, and 0≤c≤14. There are (14+1)2=225 possible combinations of values for b and c, and for every combination except (b,c)=(0,0), there is a unique a with a+b+c=28. Thus the coefficient of x28 is \boxed
OR
Let P(x)=(1+x+x2+⋯+x14)2=1+r1x+r2x2+⋯+r28x28 and Q(x)=1+x+x2+⋯+x27. The coefficient of x28 in the product P(x)Q(x) is r1+r2+⋯+r28=P(1)−1=152−1=224.
The problems on this page are the property of the MAA's American Mathematics Competitions