Determine the sum of the following series:

$$
s_{n}=5^{2}+2 \cdot 5^{2} \cdot 2 \cdot 3+3 \cdot 5^{2} \cdot 2^{2} \cdot 3^{2}+\ldots+n \cdot 5^{2} \cdot 2^{n-1} \cdot 3^{n-1}
$$
I. solution: The sum can also be written as:

$$
\begin{array}{r}
s_{n}=5^{2}\left(1+6+6^{2}+\ldots+6^{n-2}+6^{n-1}\right)+ \\
+5^{2}\left(6+6^{2}+\ldots+6^{n-2}+6^{n-1}\right)+ \\
+5^{2}\left(6^{2}+\ldots+6^{n-2}+6^{n-1}\right)+ \\
\ldots \ldots \ldots \ldots \cdots \cdots \cdots \cdots \cdots \\
\ldots \ldots \ldots \cdots \cdots \cdots \cdots \cdots \\
+5^{2}\left(6^{n-2}+6^{n-1}\right)+ \\
+5^{2} \cdot 6^{n-1}
\end{array}
$$

Each parenthesis contains a geometric series.

The $k$-th row can be simplified as follows:

$$
\begin{gathered}
s_{k}^{\prime}=5^{2}\left(6^{k-1}+6^{k}+\ldots+6^{n-1}\right)=5^{2} \cdot 6^{k-1} \frac{6^{n-k+1}-1}{6-1}= \\
=5\left(6^{n}-6^{k-1}\right)=5 \cdot 6^{n}-5 \cdot 6^{k-1}
\end{gathered}
$$

Therefore,

$$
\begin{aligned}
n=s_{1}^{\prime}+ & s_{2}^{\prime}+\ldots+s_{k}^{\prime}+\ldots+s_{n}^{\prime}=\left(5 \cdot 6^{n}-5 \cdot 6^{0}+\left(5 \cdot 6^{n}-5 \cdot 6\right)+\ldots+\right. \\
+ & \left(5 \cdot 6^{n}-5 \cdot 6^{n-1}\right)=n \cdot 5 \cdot 6^{n}-5\left(1+6+\ldots+6^{n-1}\right)= \\
& =5 n 6^{n}-5 \frac{6^{n}-1}{6-1}=5 n 6^{n}-6^{n}+1=(5 n-1) 6^{n}+1
\end{aligned}
$$

Rázga Tamás (Bp. II., Rákóczi g. III. o. t.)

II. solution: Multiplying both sides by 6 and factoring out $5^{2}$ on the right side,

$$
6 s_{n}=5^{2}\left(6+2 \cdot 6^{2}+3 \cdot 6^{2}+\ldots+(n-1) 6^{n-1}+n 6^{n}\right)
$$

Subtracting $s_{n}$ from both sides,

$$
\begin{gathered}
5 s_{n}=5^{2}\left[n 6^{n}-\left(1+6+6^{2}+\ldots+6^{n-1}\right)\right]= \\
=5^{2}\left[n 6^{n}-\frac{6^{n}-1}{6-1}\right]=5^{2} n 6^{n}-5 \cdot 6^{n}+5
\end{gathered}
$$

Dividing both sides by 5,

$$
s_{n}=5 n 6^{n}-6^{n}+1=(5 n-1) 6^{n}+1
$$

Kocsis János (Eger, Dobó István g. III. o. t.)

III. solution: If we have already figured out the relationship

$$
s_{n}=(5 n-1) 6^{n}+1
$$

for $n=1$ and $n=2$ (which can be verified by simple calculation), then we can prove the correctness of (1) for all $n$ using mathematical induction.

Assume that (1) holds for $n=k$, i.e.,

$$
s_{k}=(5 k-1) 6^{k}+1
$$

is true. We will prove that in this case (1) also holds for $n=(k+1)$. Indeed,

$$
\begin{gathered}
s_{k+1}=s_{k}+(k+1) 5^{2} \cdot 2^{k} \cdot 3^{k}=(5 k-1) 6^{k}+1+6^{k} \cdot 5^{2}(k+1)= \\
=6^{k}(5 k-1+25 k+25)+1=6^{k}(30 k+24)+1=6^{k+1}(5 k+4)+1= \\
=[5(k+1)-1] 6^{k+1}+1
\end{gathered}
$$

which is precisely (1) when $n$ is replaced by $(k+1)$.

This proves that (1) holds for any natural number $n$.