Question

The coefficient of ${{x}^{17}}$ in the expansion of $\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)$ is
\[\begin{align}
  & \text{A}.-\text{171} \\
 & \text{B}.\text{ 171} \\
 & \text{C}.\text{ 153} \\
 & \text{D}.-\text{153} \\
\end{align}\]

Answer 1

Hint: To solve this question, we will first consider $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$ and see what is highest power of x in this and then observe similarly that what will be coefficient of x in $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$ that can be obtained by putting x = 0 and adding terms of $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$. Similarly, we can calculate coefficient of ${{x}^{2}}$ in term $\left( \text{x}-\text{1} \right)\left( \text{x}-\text{2} \right)\left( x-3 \right)$. Finally, following the same process we will calculate coefficient of ${{x}^{17}}$ in \[\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)\]

Complete step-by-step answer:
Given, we have \[\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)\]
To have a better understanding of the solution, first consider; $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$ then expanding $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$ by opening the bracket, we get:
\[\begin{align}
  & \left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)=x\left( x-2 \right)-1\left( x-2 \right) \\
 & \Rightarrow {{x}^{2}}-2x-x+2 \\
 & \Rightarrow {{x}^{2}}-3x+2\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
So, we observe that highest power of x in $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$ is 2 and ${{x}^{1}}$ can be obtained by putting x = 0 in $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$ and adding them \[\text{Coefficient of }{{x}^{1}}=\left( -1 \right)+\left( -2 \right)=-3\] which is correct.
As we had two terms of $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)$ multiplied, so highest power of x is 2.
Now, consider $\left( \text{x}-\text{1} \right)\left( \text{x}-\text{2} \right)\left( x-3 \right)$
As the value of $\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)={{x}^{2}}-3x+2$ by equation (i) then substituting this value in above we get:
\[\left( \text{x}-\text{1} \right)\text{ }\left( \text{x}-\text{2} \right)\left( x-3 \right)=\left( {{x}^{2}}-3x+2 \right)\left( x-3 \right)\]
Opening the bracket to solve further, we get:
\[\begin{align}
  & \left( {{x}^{2}}-3x+2 \right)\left( x-3 \right)=x\left( {{x}^{2}}-3x+2 \right)-3\left( {{x}^{2}}-3x+2 \right) \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}+2x-3{{x}^{2}}+9x-6 \\
 & \Rightarrow {{x}^{3}}-6{{x}^{2}}+11x-6 \\
\end{align}\]
Therefore, the highest power of x in the expression $\left( \text{x}-\text{1} \right)\left( \text{x}-\text{2} \right)\left( x-3 \right)$ is 3.
This was so as we had three terms $\left( \text{x}-\text{1} \right),\left( \text{x}-\text{2} \right)\text{ and }\left( x-3 \right)$ multiplied, then the coefficient of ${{x}^{2}}$ would be obtained by putting x = 0 in $\left( \text{x}-\text{1} \right)\left( \text{x}-\text{2} \right)\left( x-3 \right)$ and adding them.
Similarly, if we multiply $\left( \text{x}-\text{1} \right)\left( \text{x}-\text{2} \right)\left( x-3 \right)\left( x-4 \right)$ then highest power of x would be 4 and coefficient of ${{x}^{3}}$ can be obtained by putting x = 0 in $\left( \text{x}-\text{1} \right)\left( \text{x}-\text{2} \right)\left( x-3 \right)$ and adding them.
Similarly, if we multiply $\left( \text{x}-\text{1} \right)\left( \text{x}-\text{2} \right)\left( x-3 \right)\left( x-4 \right)\left( x-5 \right)$ then the highest power of x would be 5.
To get the coefficient of ${{x}^{17}}$ we need the adding of terms of $\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)$ and that too by putting x = 0.
Also, as we only need coefficient of ${{x}^{17}}$ and not the value of ${{x}^{17}}$ then we can easily obtain this by putting x = 0 in the expression and adding them we get, exact coefficient of ${{x}^{17}}$
\[\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)\]
Therefore, substituting x = 0 in above and adding all, we get:
\[\begin{align}
  & \left( x-1 \right)+\left( x-2 \right)+\left( x-3 \right)+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+\left( x-18 \right) \\
 & \left( -1 \right)+\left( -2 \right)+\left( -3 \right)+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+\left( -18 \right) \\
 & \left( -1 \right)-2-3-4-\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }-18 \\
 & -\left( 1+2+3+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+18 \right) \\
\end{align}\]
This is obtained by taking (-1) common.
Now, we have a formula of sum of n term which is given as:
\[1+2+3+4+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+n=\dfrac{n\left( n+1 \right)}{2}\]
Substituting n = 18 in above formula to obtain $-\left( 1+2+3+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+18 \right)$ we get:
\[\begin{align}
  & -\left( 1+2+3+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+18 \right)=-\left( \dfrac{18\left( 18+1 \right)}{2} \right) \\
 & \Rightarrow -9\left( 19 \right) \\
 & \Rightarrow -171 \\
\end{align}\]
Therefore, the coefficient of ${{x}^{17}}$ in the expansion of $\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)$ is -171.

So, the correct answer is “Option A”.

Note: Another method to solve this question is:
We are given $\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)$ then as this equation has maximum power as 18, then sum of roots can be obtained by putting \[\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( x-18 \right)=0\]
Therefore, roots are \[x=1,2,3,4,5,6,7,8,9,........18\]
Hence, there are 18 roots.
When equation is of the form $a{{x}^{2}}+bx+c$ then formula of sum of roots is \[\Rightarrow \dfrac{-b}{a}=\dfrac{-\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{2}}}\]
Here, we have 18 degree equation, then sum of roots
\[\Rightarrow \dfrac{-\text{coefficient of }{{\text{x}}^{17}}}{\text{coefficient of }{{\text{x}}^{18}}}\]
Clearly, the coefficient of ${{x}^{18}}$ would be 1 as all variable x has coefficient 1.
\[1+2+3+\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+18=\dfrac{-\text{coefficient of }{{\text{x}}^{17}}}{1}\]
Using $1+2+\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }+n=\dfrac{n\left( n+1 \right)}{2}$ in above we get:
\[\begin{align}
  & -\text{coefficient of }{{\text{x}}^{17}}=\dfrac{18\left( 18+1 \right)}{2} \\
 & \text{coefficient of }{{\text{x}}^{17}}=-9\times 17 \\
 & \text{coefficient of }{{\text{x}}^{17}}=-171 \\
\end{align}\]
Hence, the coefficient of ${{x}^{17}}$ is -171 which is option A.