Let's write the following sum in the form of a monomial expression:『

$$
\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\ldots+\frac{n+2}{n!+(n+1)!+(n+2)!}
$$[^0]


[^0]:    ${ }^{1}$ The $n!$ symbol abbreviates the product 1. 2. 3... $(n-1) n$.
I. solution. Let us denote the required sum by $S_{n}$. For $n=1,2,3$, and 4, we have

$$
S_{1}=\frac{1}{3}, \quad S_{2}=\frac{11}{24}, \quad S_{3}=\frac{59}{120}, \quad S_{4}=\frac{359}{720}
$$

In the denominators of the last three values, 4!, 5!, and 6! appear, respectively, and their numerators are one less than half of the denominator. This regularity also holds for $S_{1}$ in the form of $2 / 6$. Therefore, these sums are of the form

$$
S_{n}=\frac{\frac{(n+2)!}{2}-1}{(n+2)!}=\frac{1}{2}-\frac{1}{(n+2)!}
$$

We will prove by induction that this relationship holds for all natural numbers $n$.

As a preparatory step, we simplify the $k$-th term of $S_{n}$, where $1 \leqq k \leqq n$. By factoring out $k!$ in the denominator, then further transforming and simplifying the fraction by $k+2$,

$$
\begin{gathered}
\frac{k+2}{k!+(k+1)!+(k+2)!}= \\
\frac{(k+2)}{k!(1+k+1+(k+1)(k+2))}=\frac{k+2}{k!(k+2)^{2}}=\frac{1}{k!(k+2)}
\end{gathered}
$$

Assume that (1) holds for some index $n$, we will show that it also holds for the index one greater. Using (2),

$$
\begin{aligned}
S_{n+1} & =S_{n}+\frac{1}{(n+1)!(n+3)}=\frac{1}{2}-\left(\frac{1}{(n+2)!}-\frac{1}{(n+1)!(n+3)}\right)= \\
& =\frac{1}{2}-\frac{n+3-(n+2)}{(n+3)!}=\frac{1}{2}-\frac{1}{(n+3)!}=\frac{1}{2}-\frac{1}{((n+1)+2)!}
\end{aligned}
$$

This follows from (1) by substituting $n+1$ for $n$. Thus, (1) is indeed true for all natural numbers. In a one-term expression form,

$$
S_{n}=\frac{(n+2)!-2}{2(n+2)!}
$$

Szeidl László (Budapest, Apáczai Csere J. high school III. grade)

II. solution. The above (2) expression can also be written as:

$$
\begin{gathered}
\frac{k+2}{k!+(k+1)!+(k+2)!}=\frac{1}{k!(k+2)}=\frac{k+1}{(k+2)!}=\frac{k+2-1}{(k+2)!}= \\
=\frac{1}{(k+1)!}-\frac{1}{(k+2)!}
\end{gathered}
$$

Transforming each term of $S_{n}$ in this way, we arrive at the second form of (1):

$$
\begin{aligned}
S_{n}=( & \left.\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\ldots+\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)+ \\
& +\left(\frac{1}{(n+1)!}-\frac{1}{(n+2)!}\right)=\frac{1}{2!}-\frac{1}{(n+2)!}
\end{aligned}
$$

Hernádi Ágnes (Budapest, Berzsenyi D. high school I. grade)

Remark. Essentially, we arrive at the same solution if we add the consecutive terms of the sum

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

to each term of $S_{n}$ and notice that the sum, based on (2),

$$
\begin{gathered}
\frac{1}{k!(k+2)}+\frac{1}{(k+2)!}=\frac{1}{(k+1)!}, \quad \text { so the total sum is } \\
\frac{1}{2!}+S_{n}^{*}-\frac{1}{(n+2)!}
\end{gathered}
$$