Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
n\left( {2n + 1} \right).\left( {3n + 1} \right) = n.\left( {6{n^2} + 5n + 1} \right) = 6{n^3} + 5{n^2} + n\\
{1^3} + {2^3} + .... + {n^3} = {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}\\
{1^2} + {2^2} + {3^2} + ..... + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\
\Rightarrow S = 1.3.4 + 2.5.7 + ...... + n.\left( {2n + 1} \right)\left( {3n + 1} \right)\\
= 6.\left( {{1^3} + {2^3} + .... + {n^3}} \right) + 5.\left( {{1^2} + {2^2} + ..... + {n^2}} \right) + \left( {1 + 2 + 3 + .... + n} \right)\\
= 6.{\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2} + 5.\frac{{n.\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2}\\
= \frac{{n\left( {n + 1} \right)}}{2}.\left[ {6.\frac{{n\left( {n + 1} \right)}}{2} + \frac{5}{3}.\left( {2n + 1} \right) + 1} \right]\\
= \frac{{n\left( {n + 1} \right)}}{2}.\left[ {3{n^2} + 3n + \frac{{10}}{3}n + \frac{8}{3}} \right]\\
= \frac{{n\left( {n + 1} \right)}}{2}\left[ {3{n^2} + \frac{{19}}{3}n + \frac{8}{3}} \right]\\
= \frac{{n\left( {n + 1} \right)\left( {9{n^2} + 19n + 8} \right)}}{6}
\end{array}\)
