Докажите равенство 1⋅22+2⋅32+⋯+(n−1)n2=n(n2−1)(3n+2)12.1 \cdot 2^2 + 2 \cdot 3^2 + \dots + \left( {n - 1} \right)n^2 = \frac{{n\left( {n^2 - 1} \right)\left( {3n + 2} \right)}} {{12}}.1⋅22+2⋅32+⋯+(n−1)n2=12n(n2−1)(3n+2).