Progression (Sequence and Series)HardBloom L3

Question

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

Options

A.8
B.27
C.112
D.2017

Solution

Sol. [Key Concept: Factor the sum $S$ to test divisibility.]

$$S = 3(1 + 2 + 3 + 4 + \ldots + 2016) = 2^{4} \times 3^{3} \times 7 \times 2017$$

$$\left\lbrack \because 6^{2} + 5^{2} + 4^{2} - 3^{2} - 2^{2} - 1^{2} = 3\lbrack(6 + 3) + (5 + 2) + (4 + 1)\rbrack \right.\ $$

$$= 3(1 + 2 + 3 + 4 + 5 + 6)\rbrack$$

✔ (a) $S{= 2}^{4} \cdot 3^{3} \cdot 7 \cdot 2017$, divisible by 8

✔ (b) divisible by 27

✔ (c) divisible by 112

✔ (d) divisible by 2017

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