## Task 2.

Calculate the sum

$$
\frac{1^{2}+2^{2}}{1 \cdot 2}+\frac{2^{2}+3^{2}}{2 \cdot 3}+\frac{3^{2}+4^{2}}{3 \cdot 4}+\cdots+\frac{99^{2}+100^{2}}{99 \cdot 100}
$$
## Solution.

Let's consider the sums of the first few terms:

$$
\begin{gathered}
\frac{1^{2}+2^{2}}{1 \cdot 2}=\frac{5}{2}=\frac{1 \cdot 5}{2} \\
\frac{1^{2}+2^{2}}{1 \cdot 2}+\frac{2^{2}+3^{2}}{2 \cdot 3}=\frac{14}{3}=\frac{2 \cdot 7}{3} \\
\frac{1^{2}+2^{2}}{1 \cdot 2}+\frac{2^{2}+3^{2}}{2 \cdot 3}+\frac{3^{2}+4^{2}}{3 \cdot 4}=\frac{27}{4}=\frac{3 \cdot 9}{4}
\end{gathered}
$$

We conjecture that for a natural number $n$, it might hold that

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

If this is correct, then the desired answer for $n=100$ is $\frac{99 \cdot 201}{100}$.

By making a conjecture, we cannot be certain of our answer—several examples do not constitute a proof. The statement that the sum of $n$ terms is $\frac{(n-1)(2 n+1)}{n}$ can be called $T(n)$.

We have already verified that the statement holds for $n=1$. To complete the proof, it is sufficient to show that the statement $T(n)$ implies $T(n+1)$. Then we are sure that the statement holds for $n=2,3, \ldots$, up to $n=100$.

We need to verify that under the assumption that $T(n)$ holds, the sum of $n+1$ terms is $\frac{n(2 n+3)}{n+1}$. Under the assumption $T(n)$, the sum of the first $n+1$ terms is

$$
\frac{(n-1)(2 n+1)}{n}+\frac{n^{2}+(n+1)^{2}}{n(n+1)}
$$

By bringing to a common denominator and simplifying, we get

$$
\begin{aligned}
\frac{(n-1)(2 n+1) \cdot(n+1)+n^{2}+(n+1)^{2}}{n(n+1)} & =\frac{2 n^{3}-2 n+n^{2}-1+n^{2}+n^{2}+2 n+1}{n(n+1)} \\
& =\frac{2 n^{3}+3 n^{2}}{n(n+1)}=\frac{n(2 n+3)}{n+1}
\end{aligned}
$$

This completes the proof. This method of proof is called mathematical induction.