Simplify the following sum:

$$
3+2 \cdot 3^{2}+3 \cdot 3^{3}+\cdots+100 \cdot 3^{100}
$$
I. Solution. Consider the following table:

$$
33^{2} 3^{3} 3^{4} \ldots 3^{99} 3^{100} 3^{2} 3^{3} 3^{4} \ldots 3^{99} 3^{100} 3^{3} 3^{4} \ldots 3^{99} 3^{100} 3^{4} \ldots 3^{99} 3^{100} \cdot \cdot 3^{99} 3^{100} 3^{100}
$$

If we add up the numbers in each column, we get exactly the terms of (1). The task is therefore to determine the sum of the numbers in the table.

Let's add the numbers row by row first, then sum the row sums! This is advantageous because in each row, the numbers form a geometric sequence. The sum of the numbers in the $k$-th row, according to the formula for the sum of a geometric sequence, is:

$$
3^{k}+3^{k+1}+\cdots+3^{100}=3^{k} \frac{3^{101-k}-1}{3-1}=\frac{3^{101}-3^{k}}{2}
$$

The sum of the row sums is - again using the formula for the sum of a geometric sequence -

$$
\begin{gathered}
\frac{3^{101}-3^{1}}{2}+\frac{3^{101}-3^{2}}{2}+\cdots+\frac{3^{101}-3^{100}}{2}=50 \cdot 3^{101}-\frac{3}{2}\left(1+3+3^{2}+\cdots+3^{99}\right)= \\
=50 \cdot 3^{101}-\frac{3}{2} \cdot \frac{3^{100}-1}{3-1}=\frac{199 \cdot 3^{101}+3}{4}
\end{gathered}
$$

II. Solution. Consider the polynomial $f(x)=1+x+x^{2}+x^{3}+\cdots+x^{100}$. It is easy to verify that

$$
x \cdot f^{\prime}(x)=x+2 x^{2}+3 x^{3}+\cdots+100 x^{100}
$$

so the task is to determine the value of $3 \cdot f^{\prime}(3)$. ( $f^{\prime}$ is the derivative of the polynomial $f$.)

According to the formula for the sum of a geometric sequence, for $x \neq 1$,

$$
f(x)=\frac{x^{101}-1}{x-1}
$$

and

$$
x f^{\prime}(x)=x \frac{101 x^{100}(x-1)-\left(x^{101}-1\right)}{(x-1)^{2}}=\frac{100 x^{102}-101 x^{101}+x}{(x-1)^{2}}
$$

Substituting $x=3$:

$$
3+2 \cdot 3^{2}+3 \cdot 3^{3}+\cdots+100 \cdot 3^{100}=\frac{100 \cdot 3^{102}-101 \cdot 3^{101}+3}{4}=\frac{199 \cdot 3^{101}+3}{4}
$$

Remark. Using both methods, it can be proven that if $n$ is a positive integer and $q \neq 1$, then

$$
q+2 \cdot q^{2}+3 \cdot q^{3}+\cdots+n \cdot q^{n}=q \frac{n q^{n+1}-(n+1) q^{n}+1}{(q-1)^{2}}
$$