Problem C. 1485. (May 2018)
C. 1485. Let \(\displaystyle x=1^2+3^2+5^2+\cdots+2017^2\) and \(\displaystyle y=2^2+4^2+6^2+\cdots+2018^2\). Evaluate the fraction
\(\displaystyle \frac{y-x}{y+x-(1\cdot 2+2\cdot3+3\cdot4+\cdots+2017\cdot 2018)}. \)
(5 pont)
Deadline expired on June 11, 2018.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Az \(\displaystyle y\) összegben a hatványalapokat felezve sorszámként szolgálhatnak. Így könnyen látható, hogy az \(\displaystyle x\) és \(\displaystyle y\) összegben is \(\displaystyle 1009\) tag van. Legyen \(\displaystyle p=1\cdot2+2\cdot3+3\cdot4+...+2017\cdot2018\).
\(\displaystyle y-x=2^2-1^2+4^2-3^2+6^2-5^2+...+2018^2-2017^2=\)
\(\displaystyle =(2+1)\cdot1+(4+3)\cdot1+(6+5)\cdot1+...+(2018+2017)\cdot1=\)
\(\displaystyle =3+7+11+...+4035=\frac{3+4035}{2}\cdot1009=2019\cdot1009.\)
\(\displaystyle y+x=1^2+2^2+3^2+4^2+...+2017^2+2018^2=\)
\(\displaystyle =1\cdot(2-1)+2\cdot(3-1)+3\cdot(4-1)+…+2018\cdot(2019-1)=\)
\(\displaystyle =1\cdot2+2\cdot3+3\cdot4+…+2018\cdot2019-(1+2+3+...+2018)=\)
\(\displaystyle =p+2018\cdot2019-2019\cdot1009.\)
\(\displaystyle A=\frac{2019\cdot1009}{p+2018\cdot2019-2019\cdot1009-p}=\frac{2019\cdot1009}{2019\cdot(2018-1009)}=1.\)
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Problems in Mathematics of KöMaL, May 2018