Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# 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