Let's determine the sum of the following series:
a) $\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{3^{3}}+\ldots+\frac{n}{2^{n}}$.
b) $1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+\ldots+n(n+1)(n+2)$.
a) I. solution: $S_{1}=\frac{1}{2}=2-\frac{2+1}{2}, S_{2}=\frac{1}{2}+\frac{2}{2^{2}}=2-\frac{2+2}{2^{2}}$, so the formula $S_{n}=2-\frac{2+n}{2^{n}}$ is correct for $n=1$ and $n=2$. We will prove the general correctness of the formula by complete induction. Suppose that for some natural number $k$ the formula is true, that is,

$$
S_{k}=2-\frac{2+k}{2^{k}}
$$

then

$$
S_{k+1}=S_{k}+\frac{k+1}{2^{k+1}}=2-\frac{2+k}{2^{k}}+\frac{k+1}{2^{k+1}}=2-\frac{4+2 k-k-1}{2^{k+1}}=2-\frac{2+(k+1)}{2^{k+1}}
$$

Thus, if the formula is true for $k$, it is also true for $(k+1)$. Since we have seen that it is true for 1 (and 2), it follows that the formula is true for all natural numbers.

Weisz Katalin (Bp., I., Szilágyi Erzsébet lg. IV. o. t.)

II. solution: $S_{n}$ can be written in the following form

$$
\begin{aligned}
S_{n} & =\frac{1}{2}+ \\
& +\frac{1}{2^{2}}+\frac{1}{2^{2}}+ \\
& +\frac{1}{2^{3}}+\frac{1}{2^{3}}+\frac{1}{2^{3}}+ \\
& \vdots \\
& +\frac{1}{2^{k}}+\frac{1}{2^{k}}+\frac{1}{2^{k}}+\ldots+\frac{1}{2^{k}}+ \\
& \vdots \\
& +\frac{1}{2^{n}}+\frac{1}{2^{n}}+\frac{1}{2^{n}}+\ldots+\frac{1}{2^{n}}+\ldots+\frac{1}{2^{n}}
\end{aligned}
$$

Denoting the sum of the terms under each other by $s_{1}, s_{2}, \ldots, s_{k}, \ldots, s_{n}$, $s_{k}$ is the sum of a geometric series whose first term is $\frac{1}{2^{k}}$, ratio is $\frac{1}{2}$, and the number of terms is $n-k+1$, that is,

$$
S_{k}=\frac{1}{2^{k}} \frac{1-\left(\frac{1}{2}\right)^{n-k+1}}{1-\frac{1}{2}}=\left(\frac{1}{2}\right)^{k-1}-\left(\frac{1}{2}\right)^{n}
$$

Thus,

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

Behringer Tibor (Bp., III., Árpád g. I. o. t.)

III. solution:

$$
S_{n}=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{3^{3}}+\ldots+\frac{n}{2^{n}}
$$

Multiplying both sides by $\frac{1}{2}$:

$$
\frac{S_{n}}{2}=\frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\ldots+\frac{n-1}{2^{n}}+\frac{n}{2^{n+1}}
$$

Subtracting (2) from (1):

$$
\frac{S_{n}}{2}=\left(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots+\frac{1}{2^{n}}\right)-\frac{n}{2^{n+1}}
$$

where the terms in parentheses form a geometric series with ratio $\frac{1}{2}$, so

$$
\frac{S_{n}}{2}=\frac{1}{2} \frac{1-\left(\frac{1}{2}\right)^{n}}{1-\frac{1}{2}}-\frac{n}{2^{n+1}}=1-\frac{1}{2^{n}}-\frac{n}{2^{n+1}}
$$

thus

$$
S_{n}=2-\frac{2+n}{2^{n}}
$$

Tömböl István (Pécs, Bányaip. mélyép. techn. III. o. t.)

b) I. solution: The method of complete induction naturally leads to the goal here as well.

$$
n=1 \text { for } S_{1}=1 \cdot 2 \cdot 3=\frac{1 \cdot 2 \cdot 3 \cdot 4}{4}=\frac{n(n+1)(n+2)(n+3)}{4}
$$

Suppose that

$$
S_{k}=\frac{k(k+1)(k+2)(k+3)}{4} \text { is true }
$$

then

$$
\begin{gathered}
S_{k+1}=S_{k}+(k+1)(k+2)(k+3)= \\
=\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)}{4}=\frac{(k+1)(k+2)(k+3)(k+4)}{4} .
\end{gathered}
$$

Thus, the formula $S_{n}=\frac{n(n+1)(n+2)(n+3)}{4}$ is indeed true for all natural $n$.

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

II. solution: Consider the sum of the following series:

$$
S_{n}=1 \cdot 2 \cdot 3 \cdot 4+2 \cdot 3 \cdot 4 \cdot 5+\ldots+n(n+1)(n+2)(n+3)
$$

Subtract this sum from itself, but in such a way that the $k$-th term of the subtrahend is subtracted from the $(k+1)$-th term of the minuend:

$$
\begin{gathered}
S_{n}-S_{n}=0=1 \cdot 2 \cdot 3 \cdot 4+2 \cdot 3 \cdot 4 \cdot(5-1)+3 \cdot 4 \cdot 5(6-2)+ \\
+\ldots+n(n+1)(n+2)[(n+3)-(n-1)]-n(n+1)(n+2)(n+3) \\
=4 \cdot S_{n}-n(n+1)(n+2)(n+3)
\end{gathered}
$$

thus

$$
S_{n}=\frac{n(n+1)(n+2)(n+3)}{4}
$$

Kálmán György (Szolnok, Beloiannisz g. III. o. t.)

III. solution: The $k$-th term of the series is $k(k+1)(k+2)=k^{3}+3 k^{2}+2 k$. The series can thus be rearranged into the following form:

$$
S_{n}=\left(1^{3}+2^{3}+\ldots+n^{3}\right)+3\left(1^{2}+2^{2}+\ldots+n^{2}\right)+2(1+2 \ldots+n)
$$

The formulas for the sum of cubes, squares, and natural numbers in parentheses are known, using them

$$
S_{n}=\left[\frac{n(n+1)}{2}\right]^{2}+3 \frac{n(n+1)(2 n+1)}{6}+2 \frac{n(n+1)}{2}=\frac{n(n+1)(n+2)(n+3)}{4}
$$

Jónás József (Gyöngyös, Vak Bottyán 2 . III. o. t.)