Question

The coefficient of \[{x^{n - 2}}\] in the polynomial \[\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) \ldots \ldots \left( {x - n} \right)\] is
A \[\dfrac{{n\left( {{n^2} + 2} \right)\left( {3n + 1} \right)}}{{24}}\]
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

Answer 1

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 \[{x^{n - 2}}\].

Complete step by step answer:
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.
\[
  \,\,\,\,\,{\text{coefficient of }}{x^{n - 2}}\left( C \right) = \left( {1 \times 2 + 1 \times 3 + \ldots + 1 \times n} \right) + \left( {2 \times 3 + 2 \times 4 + \ldots + 2 \times n} \right) + \ldots + \left( {\left( {n - 1} \right) \times n} \right) \\
   \Rightarrow C = \left( {\dfrac{{n\left( {n + 1} \right)}}{2} - 1} \right) + 2\left( {\dfrac{{n\left( {n + 1} \right)}}{2} - 1 - 2} \right) + \ldots + \left( {n - 1} \right)\left( {\dfrac{{n\left( {n + 1} \right)}}{2} - \left( {1 + 2 + \ldots + \left( {n - 1} \right)} \right)} \right) \\
   \Rightarrow C = \left\{ {\left( {1 + 2 + \ldots + \left( {n - 1} \right)} \right)\dfrac{{n\left( {n + 1} \right)}}{2}} \right\} - \left\{ {1 + 2\left( {1 + 2} \right) + 3\left( {1 + 2 + 3} \right) + \ldots + \left( {n - 1} \right)\left( {1 + \ldots + \left( {n - 1} \right)} \right)} \right\} \\
   \Rightarrow C = \left\{ {\dfrac{{\left( {n - 1} \right)n}}{2} \times \dfrac{{n\left( {n + 1} \right)}}{2}} \right\} - \left\{ {\sum\limits_1^{n - 1} {k\left( {\dfrac{{k\left( {k + 1} \right)}}{2}} \right)} } \right\} \\
   \Rightarrow C = \left\{ {\dfrac{{{n^2}\left( {{n^2} - 1} \right)}}{4}} \right\} - \left\{ {\sum\limits_1^{n - 1} {\dfrac{{{k^3} + {k^2}}}{2}} } \right\} \\
\]
Further, simplify the above expression.
\[
  \,\,\,\,\,\,C = \left\{ {\dfrac{{{n^2}\left( {{n^2} - 1} \right)}}{4}} \right\} - \dfrac{1}{2}\left\{ {{{\left( {\dfrac{{\left( {n - 1} \right)n}}{2}} \right)}^2} + \dfrac{{\left( {n - 1} \right)n\left( {2n - 1} \right)}}{6}} \right\} \\
   \Rightarrow C = \dfrac{{n\left( {n - 1} \right)}}{4}\left\{ {n\left( {n + 1} \right) - \dfrac{{n\left( {n - 1} \right)}}{2} - \dfrac{{2n - 1}}{3}} \right\} \\
   \Rightarrow C = \dfrac{{n\left( {n - 1} \right)}}{4}\left\{ {\dfrac{{n\left( {n + 3} \right)}}{2} - \dfrac{{2n - 1}}{3}} \right\} \\
   \Rightarrow C = \dfrac{{n\left( {n - 1} \right)}}{4}\left\{ {\dfrac{{3{n^2} + 9n - 4n + 2}}{6}} \right\} \\
   \Rightarrow C = \dfrac{{n\left( {{n^2} - 1} \right)\left( {3n + 2} \right)}}{{24}} \\
\]
Thus, the coefficient of \[{x^{n - 2}}\] is \[\dfrac{{n\left( {{n^2} - 1} \right)\left( {3n + 2} \right)}}{{24}}\]

So, the correct answer is “Option B”.

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 \[\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\].