Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}} + \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\\
= \frac{1}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}.k\left( {k + 3} \right) + \frac{1}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}.\frac{4}{{k + 3}}\\
= \frac{1}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {k\left( {k + 3} \right) + \frac{4}{{k + 3}}} \right]\\
= \frac{1}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}\left( {\frac{{k{{\left( {k + 3} \right)}^3} + 4}}{{k + 3}}} \right)\\
= \frac{1}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}.\frac{{k\left( {{k^2} + 6k + 9} \right) + 4}}{{k + 3}}\\
= \frac{{{k^3} + 6{k^2} + 9k + 4}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\\
= \frac{{\left( {{k^3} + {k^2}} \right) + \left( {5{k^2} + 5k} \right) + \left( {4k + 4} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\\
= \frac{{\left( {k + 1} \right)\left( {{k^2} + 5k + 4} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\\
= \frac{{\left( {k + 1} \right)\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\\
= \frac{{\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}
\end{array}\)
