## Task A-4.4.

Let $n$ be a natural number. Determine all positive real numbers $x$ for which

$$
\frac{2^{2}}{x+1}+\frac{3^{2}}{x+2}+\cdots+\frac{(n+1)^{2}}{x+n}+n x^{2}=n x+\frac{n(n+3)}{2}
$$
## First Solution.

Since $n x=x+x+\cdots+x$, and $\frac{n(n+3)}{2}=(1+2+3+\cdots+n)+n$, the original equality is equivalent to

$$
\sum_{k=1}^{n} \frac{(k+1)^{2}}{x+k}+n x^{2}-\sum_{k=1}^{n} x-\sum_{k=1}^{n} k-n=0
$$

or equivalently,

$$
\sum_{k=1}^{n}\left(\frac{(k+1)^{2}}{x+k}-(x+k)\right)+n x^{2}-n=0
$$

By simplifying, we get

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

Thus, the above equality is equivalent to

$$
(1-x)\left(n+\sum_{k=1}^{n} \frac{k+1}{x+k}\right)+n\left(x^{2}-1\right)=0
$$

or equivalently,

$$
(1-x)\left(n+\left(\frac{2}{x+1}+\frac{3}{x+2}+\cdots+\frac{n+1}{x+n}\right)-n(1+x)\right)=0
$$

For $x=1$, the equality clearly holds. For $x \neq 1$, it must hold that

$$
n+\left(\frac{2}{x+1}+\frac{3}{x+2}+\cdots+\frac{n+1}{x+n}\right)=n(1+x)
$$

or equivalently,

$$
\frac{2}{x+1}+\frac{3}{x+2}+\cdots+\frac{n+1}{x+n}=n x
$$

If $0 < x < 1$, the left side is strictly less than $n$, while the right side is strictly greater than $n$. If $x > 1$, the opposite is true: the left side is strictly less than $n$, and the right side is strictly greater than $n$. Therefore, the only solution is $x=1$.