Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.9 AMC 10/12 Find the answer Ukraine

Andriy has counted the sum of the squares of positive integers from $1$ to $2016$ , Vitaliy has counted the sum of the squares of natural numbers from $1$ to $2016$ . Yuriy has added Andriy's and Vitaliy's numbers, multiplied by $3$ and added $2016$ . What number has Yuriy got?

(Andriy Anikushin)

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let's write down the number that Yuriy had got:
$\begin{aligned} & 3 \cdot (1^2 + 2^2 + \dots + 2016^2) + 3 \cdot (1 + 2 + \dots + 2016) + 2016 = \\ & = (3 \cdot 1^2 + 3 \cdot 1 + 1) + (3 \cdot 2^2 + 3 \cdot 2 + 1) + \dots + (3 \cdot 2016^2 + 3 \cdot 2016 + 1) = \\ & = (1^3 + 3 \cdot 1^2 + 3 \cdot 1 + 1) + (2^3 + 3 \cdot 2^2 + 3 \cdot 2 + 1) + \dots + (2016^3 + 3 \cdot 2016^2 + 3 \cdot 2016 + 1) - \\ & \quad - (1^3 + 2^3 + \dots + 2016^3) = ((1+1)^3 + (2+1)^3 + \dots + (2016+1)^3) \\ & \quad - (1^3 + 2^3 + \dots + 2016^3) = \\ & = (2^3 + 3^3 + \dots + 2017^3) - (1^3 + 2^3 + \dots + 2016^3) = 2017^3 - 1. \end{aligned}$

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