Problem:
The expression
(x+y+z)2006+(x−y−z)2006
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
Answer Choices:
A. 6018
B. 671,676
C. 1,007,514
D. 1,008,016
E. 2,015,028
Solution:
There is exactly one term in the simplified expression for every monomial of the form xaybzc, where a,b, and c are non-negative integers, a is even, and a+b+c=2006. There are 1004 even values of a with 0≤a≤2006. For each such value, b can assume any of the 2007−a integer values between 0 and 2006−a, inclusive, and the value of c is then uniquely determined as 2006−a−b. Thus the number of terms in the simplified expression is
(2007−0)+(2007−2)+⋯+(2007−2006)=2007+2005+⋯+1
This is the sum of the first 1004 odd positive integers, which is 10042=1,008,016.
OR
The given expression is equal to
∑a!b!c!2006!​(xaybzc+xa(−y)b(−z)c)
where the sum is taken over all non-negative integers a,b, and c with a+b+c= 2006. Because the number of non-negative integer solutions of a+b+c=k is (2k+2​), the sum is taken over (22008​) terms, but those for which b and c have opposite parity have a sum of zero. If b is odd and c is even, then a is odd, so a=2A+1,b=2B+1, and c=2C for some non-negative integers A,B, and C. Therefore 2A+1+2B+1+2C=2006, so A+B+C=1002. Because the last equation has (21004​) non-negative integer solutions, there are (21004​) terms for which b is odd and c is even. The number of terms for which b is even and c is odd is the same. Thus the number of terms in the simplified expression is
(22008​)−2(21004​)=1004⋅2007−1004⋅1003=10042=1,008,016​
The problems on this page are the property of the MAA's American Mathematics Competitions