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I understand the answer, but I am stuck on blue underline part. Please someone describe the blue underline section and how it's come from. Thank you.

Source

Number theory: Structure , Examples and Problems by Titu Andreescu .

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    $\begingroup$ $$n(n^4\!-\!5n^2\!+\!4) = \color{#c00}n(\color{#0a0}{n^2\!-\!1})(n^2\!-\!4) = (n\!+\!2)(\color{#0a0}{n\!+\!1})\color{#c00}n(\color{#0a0}{n\!-\!1})(n\!-\!2) = \dfrac{(n\!+\!2)!}{(n\!-\!3)!} = 5!{n\!+\!2\choose 5}$$ $\endgroup$ Jul 13 at 16:52

2 Answers 2

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$$5!\binom{n+2}{5} = 5!\frac{(n+2)!}{5!(n-3)!} = \frac{(n+2)!}{(n-3)!} $$$$= (n+2)(n+1)(n)(n-1)(n-2)=n^5-5n^3+4n.$$

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Since $$ \left( \begin{array}{l} n + 2 \\ \,\,\,\,5 \\ \end{array} \right) = \frac{{\left( {n + 2} \right)!}}{{\left( {n + 2 - 5} \right)!\,5!}} $$ it is $$ \begin{array}{l} 5!\left( \begin{array}{l} n + 2 \\ \,\,\,\,5 \\ \end{array} \right) = \frac{{\left( {n + 2} \right)!}}{{\left( {n - 3} \right)!\,}} = \frac{{\left( {n + 2} \right)(n + 1)n(n - 1)(n - 2)(n - 3)!}}{{(n - 3)!}} = \\ \\ = \left( {n + 2} \right)(n + 1)n(n - 1)(n - 2) \\ \end{array} $$ that is what you need.

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