Mathematical and Physical Journal
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Problem K. 296. (September 2011)

K. 296. The first term of a sequence is 2011. From the second term onwards, each term is equal to (-2) times the reciprocal of the number 2 greater than the previous term. What is the 2011th term of the sequence?

(6 pont)

Deadline expired on October 10, 2011.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A sorozat második eleme \(\displaystyle -2/2013\). A harmadik elem \(\displaystyle -4026/4024=-2013/2012\), a negyedik \(\displaystyle -4024/2011\), az ötödik \(\displaystyle 2011\), ahonnan kezdve a sorozat elemei sorban ismétlődnek: \(\displaystyle a_{n+4}=a_n\). Mivel 2011 néggyel osztva 3 maradékot ad, ezért \(\displaystyle a_{2011}=a_3=-\frac{2013}{2012}\).


208 students sent a solution.
6 points:88 students.
5 points:27 students.
4 points:20 students.
3 points:19 students.
2 points:12 students.
1 point:7 students.
0 point:32 students.
Unfair, not evaluated:3 solutions.

Problems in Mathematics of KöMaL, September 2011