Sequences & Series
Series and Sums
Grade 11
Question:
<p>Let \(S = 2016^2 + 2015^2 + 2014^2 - 2013^2 - 2012^2 - 2011^2 + 2010^2 + 2009^2 + 2008^2 - 2007^2 - 2006^2 - 2005^2 + \ldots + 6^2 + 5^2 + 4^2 - 3^2 - 2^2 - 1^2\), then \(S\) is divisible by</p>
<p>(A) 8</p>
<p>(B) 27</p>
<p>(C) 112</p>
<p>(D) 2017</p>
Step-by-Step Solution
Key Concept: Grouping consecutive terms and using algebraic identities reveals a common factor pattern.
<p>The sum can be grouped as \((2016^2 + 2015^2 + 2014^2 - 2013^2 - 2012^2 - 2011^2) + \ldots\). Each group has the form \((n^2 + (n-1)^2 + (n-2)^2 - (n-3)^2 - (n-4)^2 - (n-5)^2)\). Using the difference of squares and factoring, this pattern yields divisibility by 112.</p>
Correct Answer: C