Problem C. 1462. (February 2018)

C. 1462. The first term of an arithmetic sequence is \(\displaystyle a_1=3\), and its common difference is 9. Prove that for every natural number \(\displaystyle k\), the number \(\displaystyle 3\cdot 4^k\) occurs among the terms.

(5 pont)

Deadline expired on March 12, 2018.

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

Megoldás. A sorozat \(\displaystyle n\)-edik eleme: \(\displaystyle a_n=3+(n-1)\cdot9=3+9m=3(1+3m)\). Bizonyítandó, hogy tetszőleges természetes \(\displaystyle k\) számra van olyan \(\displaystyle m\) pozitív egész, melyre \(\displaystyle 3(1+3m)=3\cdot4^k\). Az egyenlet mindkét oldalát \(\displaystyle 3\)-mal osztva és levonva \(\displaystyle 1\)-et: \(\displaystyle 3m=4^k-1\). A jobb oldali kifejezés minden \(\displaystyle k\) egész szám esetén szorzattá alakítható: \(\displaystyle 3m=4^k-1^k=(4-1)(4^{k-1}+4^{k-2}+...+4+1)\). Vagyis ha \(\displaystyle m=4^{k-1}+4^{k-2}+...+4+1\), akkor \(\displaystyle a_{9m+3}=3\cdot4^k\).

Statistics:

100 students sent a solution.
5 points:	57 students.
4 points:	18 students.
3 points:	11 students.
2 points:	3 students.
1 point:	8 students.
0 point:	3 students.

Problems in Mathematics of KöMaL, February 2018