Chứng minh rằng với mọi n∈N*: 1²+2²+…+n²=n(n+1)(2n+1)/6

Giải thích các bước giải:

 Ta có:

\[\begin{array}{l}
{1^2} + {2^2} + {3^2} + {4^2} + …. + {n^2}\\
 = 1.1 + 2.2 + 3.3 + 4.4 + ….. + n.n\\
 = 1.\left( {2 – 1} \right) + 2\left( {3 – 1} \right) + 3.\left( {4 – 1} \right) + …. + n\left[ {\left( {n + 1} \right) – 1} \right]\\
 = 1.2 – 1. + 2.3 – 2 + 3.4 – 3 + 4.5 – 4 + …. + n\left( {n + 1} \right) – n\\
 = \left( {1.2 + 2.3 + 3.4 + 4.5 + ….. + n\left( {n + 1} \right)} \right) – \left( {1 + 2 + 3 + 4 + …. + n} \right)\\
 = \frac{1}{3}.\left[ \begin{array}{l}
1.2.\left( {3 – 0} \right) + 2.3.\left( {4 – 1} \right) + 3.4.\left( {5 – 2} \right) + ….\\
 + n\left( {n + 1} \right)\left( {\left( {n + 2} \right) – \left( {n – 1} \right)} \right)
\end{array} \right] – \frac{{n\left( {n + 1} \right)}}{2}\\
 = \frac{1}{3}\left[ \begin{array}{l}
1.2.3 – 0.1.2 + 2.3.4 – 1.2.3 + 3.4.5 – 2.3.4 + …..\\
 + n\left( {n + 1} \right)\left( {n + 2} \right) – \left( {n – 1} \right)n\left( {n + 1} \right)
\end{array} \right] – \frac{{n\left( {n + 1} \right)}}{2}\\
 = \frac{1}{3}\left[ {n\left( {n + 1} \right)\left( {n + 2} \right) – 0.1.2} \right] – \frac{{n\left( {n + 1} \right)}}{2}\\
 = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3} – \frac{{n\left( {n + 1} \right)}}{2}\\
 = n\left( {n + 1} \right)\left( {\frac{{n + 2}}{3} – \frac{1}{2}} \right)\\
 = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}
\end{array}\]