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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 points)

This problem is for grade 9 students only.

Deadline expired on 10 October 2011.

Google Translation (Sorry, the solution is published in Hungarian only.)

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 $\displaystyle a_{2011}=a_3=-\frac{2013}{2012}$.

Statistics on problem K. 296.
 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

•  Támogatóink: Morgan Stanley