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The coefficient of ${x^{n - 2}}$ in the polynomial $\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) \ldots \ldots \left( {x - n} \right)$ isA $\dfrac{{n\left( {{n^2} + 2} \right)\left( {3n + 1} \right)}}{{24}}$B $\dfrac{{n\left( {{n^2} - 1} \right)\left( {3n + 2} \right)}}{{24}}$C $\dfrac{{n\left( {{n^2} + 1} \right)\left( {3n + 4} \right)}}{{24}}$D None of these

Last updated date: 20th Sep 2024
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Hint: In this problem, first we need to choose the integers from any two brackets and $x$ from all other brackets to form the term of ${x^{n - 2}}$. Next, find the coefficient of ${x^{n - 2}}$.

From the given expression, it can be observed that there are n brackets. To form the term ${x^{n - 2}}$, we need to choose integers from any two brackets and $x$ from all the other brackets and multiplied.
Now, the coefficient of ${x^{n - 2}}$ is calculated as shown below.
$\,\,\,\,\,{\text{coefficient of }}{x^{n - 2}}\left( C \right) = \left( {1 \times 2 + 1 \times 3 + \ldots + 1 \times n} \right) + \left( {2 \times 3 + 2 \times 4 + \ldots + 2 \times n} \right) + \ldots + \left( {\left( {n - 1} \right) \times n} \right) \\ \Rightarrow C = \left( {\dfrac{{n\left( {n + 1} \right)}}{2} - 1} \right) + 2\left( {\dfrac{{n\left( {n + 1} \right)}}{2} - 1 - 2} \right) + \ldots + \left( {n - 1} \right)\left( {\dfrac{{n\left( {n + 1} \right)}}{2} - \left( {1 + 2 + \ldots + \left( {n - 1} \right)} \right)} \right) \\ \Rightarrow C = \left\{ {\left( {1 + 2 + \ldots + \left( {n - 1} \right)} \right)\dfrac{{n\left( {n + 1} \right)}}{2}} \right\} - \left\{ {1 + 2\left( {1 + 2} \right) + 3\left( {1 + 2 + 3} \right) + \ldots + \left( {n - 1} \right)\left( {1 + \ldots + \left( {n - 1} \right)} \right)} \right\} \\ \Rightarrow C = \left\{ {\dfrac{{\left( {n - 1} \right)n}}{2} \times \dfrac{{n\left( {n + 1} \right)}}{2}} \right\} - \left\{ {\sum\limits_1^{n - 1} {k\left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right)} } \right\} \\ \Rightarrow C = \left\{ {\dfrac{{{n^2}\left( {{n^2} - 1} \right)}}{4}} \right\} - \left\{ {\sum\limits_1^{n - 1} {\dfrac{{{k^3} + {k^2}}}{2}} } \right\} \\$
$\,\,\,\,\,\,C = \left\{ {\dfrac{{{n^2}\left( {{n^2} - 1} \right)}}{4}} \right\} - \dfrac{1}{2}\left\{ {{{\left( {\dfrac{{\left( {n - 1} \right)n}}{2}} \right)}^2} + \dfrac{{\left( {n - 1} \right)n\left( {2n - 1} \right)}}{6}} \right\} \\ \Rightarrow C = \dfrac{{n\left( {n - 1} \right)}}{4}\left\{ {n\left( {n + 1} \right) - \dfrac{{n\left( {n - 1} \right)}}{2} - \dfrac{{2n - 1}}{3}} \right\} \\ \Rightarrow C = \dfrac{{n\left( {n - 1} \right)}}{4}\left\{ {\dfrac{{n\left( {n + 3} \right)}}{2} - \dfrac{{2n - 1}}{3}} \right\} \\ \Rightarrow C = \dfrac{{n\left( {n - 1} \right)}}{4}\left\{ {\dfrac{{3{n^2} + 9n - 4n + 2}}{6}} \right\} \\ \Rightarrow C = \dfrac{{n\left( {{n^2} - 1} \right)\left( {3n + 2} \right)}}{{24}} \\$
Thus, the coefficient of ${x^{n - 2}}$ is $\dfrac{{n\left( {{n^2} - 1} \right)\left( {3n + 2} \right)}}{{24}}$
Note: The formula for the sum of $n$ natural number is $\dfrac{{n\left( {n + 1} \right)}}{2}$. The formula for the sum of the square of the $n$ natural number is $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$.