Problem:
The polynomial 1−x+x2−x3+⋯+x16−x17 may be written in the form a0+a1y+a2y2+a3y3+⋯+a16y16+a17y17, where y=x+1 and the ai 's are constants. Find the value of a2.
Solution:
Substituting y−1 for x, the given expression becomes
1−(y−1)+(y−1)2−(y−1)3+⋯+(y−1)16−(y−1)17
which may be written in the form
1+(1−y)+(1−y)2+(1−y)3+⋯+(1−y)16+(1−y)17(1)
Note that each (1−y)k term in (1) will yield a y2 term for 2≤k≤17. More specifically, by the Binomial Theorem, each of the summands in (1) contributes (2k) or 2k(k−1) to the coefficient of y2. Therefore, the problem is equivalent to computing the sum
(22)+(23)+(24)+⋯+(217)
To this end, one may proceed directly (i.e., by calculating and adding the sixteen numbers 1,3,6,…,136) or use the result of the derivation
k=2∑n2k(k−1)=k=1∑n2k(k−1)=21(k=1∑nk2−k=1∑nk)=21(6n(n+1)(2n+1)−2n(n+1))=6(n+1)n(n−1);
or use the more general formula
k=m∑n(mk)=(m+1n+1)
In any case, the desired sum is equal to 816.
The problems on this page are the property of the MAA's American Mathematics Competitions