Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Prove it United States

Problem:
Determine, with proof, the value of
12−22+32−42+52−⋯+972−982+992. 1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-\cdots+97^{2}-98^{2}+99^{2} .

Solution

Solution:
Observe that 32−22=(3−2)(3+2)=3+23^{2}-2^{2}=(3-2)(3+2)=3+2, 52−42=(5−4)(5+4)=5+45^{2}-4^{2}=(5-4)(5+4)=5+4, and so on. Thus this sum, call it SS, is actually equal to
S=1+(2+3)+(4+5)+⋯+(97+98)+99. S=1+(2+3)+(4+5)+\cdots+(97+98)+99 .
We can also write it in reverse as
S=99+98+97+96+95+⋯+3+2+1. S=99+98+97+96+95+\cdots+3+2+1 .
Adding these two, we get
2S=100+100+⋯+100⏟99 times 2 S=\underbrace{100+100+\cdots+100}_{99 \text{ times}}
Thus, S=12⋅100⋅99=4950S=\frac{1}{2} \cdot 100 \cdot 99=4950.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.