Problem:
The polynomial 1−x+x2−x3+⋯+x16−x17 may be written in the form a0​+a1​y+a2​y2+a3​y3+⋯+a16​y16+a17​y17, 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∑n​2k(k−1)​=k=1∑n​2k(k−1)​​=21​(k=1∑n​k2−k=1∑n​k)=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