26. Calculate the sum $1^{2}+2^{2}+\ldots+n^{2}$,
26. Let $u_{n}=1+2^{2}+\ldots+(n-1)^{2}+n^{2}$. Then it is clear that $\Delta u_{n}=n^{2}$. From problem 24, it follows that $u_{n}$ can be expressed as a polynomial of the 3rd degree, i.e.,

$$
u_{n}=a_{0} n^{3}+a_{1} n^{2}+a_{2} n+a_{3}
$$

Now let's find $a_{0}, a_{1}, a_{2}, a_{3}$. For this, substitute the numbers $0,1,2,3$ for $n$, and we get

$$
\left.\begin{array}{l}
u_{0}=a_{3}=0 \\
u_{1}=a_{0}+a_{1}+a_{2}+a_{3}=1 \\
u_{2}=8 a_{0}+4 a_{1}+2 a_{2}+a_{3}=5 \\
u_{3}=27 a_{0}+9 a_{1}+3 a_{2}+a_{3}=14
\end{array}\right\}
$$

Solve the obtained system:

$$
\left.\begin{array}{r}
a_{0}+a_{1}+a_{2}=1 \\
8 a_{0}+4 a_{1}+2 a_{2}=5 \\
27 a_{0}+9 a_{1}+3 a_{2}=14
\end{array}\right\}
$$

For this, subtract three times the first equation from the third, and two times the first equation from the second. Then we get

$$
\begin{aligned}
& 24 a_{0}+6 a_{1}=11, \quad \\
& 6 a_{0}+2 a_{1}=3 \\
& 6 a_{0}=2, \quad a_{0}=\frac{1}{3}, \quad a_{1}=\frac{1}{2}, \quad a_{2}=\frac{1}{6}
\end{aligned}
$$

Thus,

$$
\begin{aligned}
& u_{n}=\frac{1}{3} n^{3}+\frac{1}{2} n^{2}+\frac{1}{6} n= \\
& \quad=\frac{n\left(2 n^{2}+3 n+1\right)}{6}=\frac{n(n+1)(2 n+1)}{6}
\end{aligned}
$$

i.e., $1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$.

Remark. We could have avoided using the results of problem 24. For this, we needed to, after obtaining the answer, prove it by mathematical induction.