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Algebra Difficulty 4.6 AIME Prove it Saudi Arabia

Find the minimum of ∑k=040(x+k2)2\sum_{k=0}^{40}\left(x+\frac{k}{2}\right)^{2} where xx is a real number.

Solution

We have
∑k=040(x+k2)2=(x+10)2+∑k=120(((x+10)+k2)2+((x+10)−k2)2)=(x+10)2+2∑k=120((x+10)2+(k2)2)=41(x+10)2+12∑k=120k2=41(x+10)2+20⋅21⋅412⋅6=41(x+10)2+1435 \begin{aligned} \sum_{k=0}^{40}\left(x+\frac{k}{2}\right)^{2} & =(x+10)^{2}+\sum_{k=1}^{20}\left(((x+10)+\frac{k}{2})^{2}+((x+10)-\frac{k}{2})^{2}\right) \\ & =(x+10)^{2}+2 \sum_{k=1}^{20}\left((x+10)^{2}+\left(\frac{k}{2}\right)^{2}\right) \\ & =41(x+10)^{2}+\frac{1}{2} \sum_{k=1}^{20} k^{2}=41(x+10)^{2}+\frac{20 \cdot 21 \cdot 41}{2 \cdot 6} \\ & =41(x+10)^{2}+1435 \end{aligned}
Therefore, the minimum value is 14351435, which is obtained when x=−10x=-10.

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