any idea?

Let us fix $n$, and let \[ P_m = \sum^m_{k=0} \binom{3m}{3k} (3n-1)^k \] we are required to show that $3^m n$ divides $P_m$ when $m$ is a positive odd integer. Now either this is a very computationally heavy problem, or I'm just not using the right methods. Let $\omega = \frac{-1 + i\sqrt{3}}{2}$ be the 3rd root of unity. Consider the coefficient of $x$ in $(1+x)^{3m} + (1+\omega x)^{3m} + (1+\omega^2 x)^{3m}$. By noting that $1 + \omega + \omega^2=0$, we see that the coefficient of $x^i$ is $0$ if $3 \nmid i$, and $\binom{3m}{i}$ if $3 \mid i$. So \[ (1+x)^{3m} + (1+\omega x)^{3m} + (1+\omega^2 x)^{3m} = 3\sum^m_{k=0} \binom{3m}{3k} x^{3k} \] so \[ P_m = \frac{A^m + B^m + C^m}{3} \] where $A= (1+ \sqrt[3]{3n-1})^3$, $B= (1+ \omega \sqrt[3]{3n-1})^3$, $C = (1+ \omega^2 \sqrt[3]{3n-1})^3$. Suppose that $m \geq 7$. Using the identity \begin{eqnarray*} A^m + B^m + C^m &=& (A^2 + B^2 + C^2)(A^{m-2} + B^{m-2} + C^{m-2}) \\ && - (A^2B^2 + B^2C^2 + C^2A^2)(A^{m-4} + B^{m-4} + C^{m-4}) \\ && + A^2B^2C^2(A^{m-6} + B^{m-6} + C^{m-6}) \end{eqnarray*} we find that \[ P_m = (A^2 + B^2 + C^2)P_{m-2} - (A^2B^2 + B^2C^2 + C^2A^2)P_{m-4} + A^2B^2C^2P_{m-6} \] Now, computing, we get \begin{eqnarray*} A^2 + B^2 + C^2 &=& 3^3(n^2+6n-2) \\ A^2B^2 + B^2C^2 + C^2A^2 &=& 3^5(n^4-18n^3+33n^2-18n+3) \\ A^2B^2C^2 &=& 3^6 n^6 \end{eqnarray*} so \[ P_m = 3^3(n^2+6n-2)P_{m-2} - 3^5(n^4-18n^3+33n^2-18n+3)P_{m-4} + 3^6 n^6 P_{m-6} \] It follows that if $P_{m-2}$, $P_{m-4}$, $P_{m-6}$ satisfy the result, then so does $P_m$. It suffices now to show the three base cases: \begin{eqnarray*} \frac{P_1}{3n} &=& 1 \\ \frac{P_3}{3^3n} &=& n^2+27n-9 \\ \frac{P_5}{3^5n} &=& n^4+150n^3+355n^2-270n+45 \end{eqnarray*} and we are done by induction.

hey, I had posted a solution for this one before and it dissapeared!

We can solve this problem by induction. Used $(1+\sqrt[3]{3n-1})^{3m}$

Although it was written in Vietnamese , I think you coud understand it clearly

Attachments:
Số Học Bài 14.pdf (219kb)

billzhao wrote: $ P_m = 3^3(n^2 + 6n - 2)P_{m - 2} - 3^5(n^4 - 18n^3 + 33n^2 - 18n + 3)P_{m - 4} + 3^6 n^6 P_{m - 6}$ Or we can use: With $ x$ denote for $ A,B,C$ we have $ (\sqrt [3]{x} - 1)^3 = 3n - 1\Rightarrow x - 3\sqrt [3]{x^2} + 3\sqrt [3]{x} - 1 = 3n - 1$ $ \Rightarrow x - 3n = 3\sqrt [3]{x}(\sqrt [3]{x} - 1)\Rightarrow (x - 3n)^3 = 27x(3n - 1)$ $ \Rightarrow x^3 - 9nx^2 + (27n^2 - 81n + 27)x - 27n^3 = 0$ $ \Rightarrow P_{m + 3} - 9nP_{m + 2} + (27n^2 - 81n + 27)P_{m + 1} - 27n^3P_{m} = 0$ Then OK by induction! PS: We also can calculate $ A + B + C,AB + BC + CA,ABC$ to have $ P_{m + 3} - 9nP_{m + 2} + (27n^2 - 81n + 27)P_{m + 1} - 27n^3P_{m} = 0$