Problem K/C. 788. (November 2023)
K/C. 788. Sequence \(\displaystyle a_n\) satisfies \(\displaystyle a_1=2\) and \(\displaystyle a_{n+1} = a_n + 2n\). Find the value of \(\displaystyle a_{100}\).
(5 pont)
Deadline expired on December 11, 2023.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás.
$$\begin{align*} a_{n+1} -a_{n} &= 2n,\\ a_{2} -a_{1} &= 2 \cdot 1,\\ a_{3} -a_{2} &= 2 \cdot 2,\\ \ldots{}\\ a_{100} -a_{99} &= 2 \cdot 99. \end{align*}$$Ezeket összeadva \(\displaystyle a_{100} -a_{99} + a_{99} -a_{98} + \ldots{} + a_{3} -a_{2} + a_{2} - a_{1} = 2 \cdot (99 + 98 + \ldots{} + 2 + 1)\), azaz \(\displaystyle a_{100} - a_{1} = 2 \cdot (1 + 99) : 2 \cdot 99 = 9900\). Tehát \(\displaystyle a_{100} = 9902\).
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Problems in Mathematics of KöMaL, November 2023