Question

If the coefficient of ${x^3}$ and ${x^4}$ in $(1 + ax + b{x^2}){(1 - 2x)^{18}}$ in powers of $x$ are both zero, then find the value of $(a,b)$ .
A. $\left( {16,\dfrac{{251}}{3}} \right)$
B. $\left( {14,\dfrac{{251}}{3}} \right)$
C. $\left( {13,\dfrac{{272}}{3}} \right)$
D. $\left( {16,\dfrac{{272}}{3}} \right)$

Answer 1

We will first open the bracket of $(1 + ax + b{x^2})$ using distributive property and then, find the coefficient of ${x^3}$ in ${(1 - 2x)^{18}}$, ${x^2}$ in $a{(1 - 2x)^{18}}$ and $x$ in $b{(1 - 2x)^{18}}$ and equate it to 0 and similarly, coefficient of ${x^4}$ in ${(1 - 2x)^{18}}$, ${x^3}$ in $a{(1 - 2x)^{18}}$ and ${x^2}$ in $b{(1 - 2x)^{18}}$ and equate it to 0. We will thus have our answer.

Complete step-by-step answer:
Let us first look at our given expression which is $(1 + ax + b{x^2}){(1 - 2x)^{18}}$.
We can rewrite the same as $(1 + ax + b{x^2}){(1 - 2x)^{18}} = {(1 - 2x)^{18}} + ax{(1 - 2x)^{18}} + b{x^2}{(1 - 2x)^{18}}$.
Now, to find the coefficient of ${x^n}$ in ${(1 - 2x)^{18}} + ax{(1 - 2x)^{18}} + b{x^2}{(1 - 2x)^{18}}$ is equivalent to finding the coefficient of ${x^n}$ in ${(1 - 2x)^{18}}$, ${x^{n - 1}}$ in $a{(1 - 2x)^{18}}$ and ${x^{n - 2}}$ in $b{(1 - 2x)^{18}}$.
Let us first discuss the coefficient of ${x^r}$ in ${(a + bx)^n}$.
Coefficient of ${x^r}$ in ${(a + bx)^n}$ is given by $\dfrac{{n!}}{{r!\left( {n - r} \right)!}} \times {a^{n - r}} \times {b^r}$. ……….(1)
Applying this approach to get the coefficients of ${x^3}$ $(1 + ax + b{x^2}){(1 - 2x)^{18}}$ by finding the coefficient of ${x^3}$ in ${(1 - 2x)^{18}}$, ${x^2}$ in $a{(1 - 2x)^{18}}$ and $x$ in $b{(1 - 2x)^{18}}$.
So, the coefficient of ${x^3}$ in ${(1 - 2x)^{18}}$ using (1) will be:- $\dfrac{{18!}}{{3!\left( {15} \right)!}} \times {1^{18 - 3}} \times {( - 2)^3}$ because on comparing, we have $a = 1,b = - 2,n = 18$ and ${\text{ }}r = 3$.
We know that $n! = n.(n - 1).(n - 2)........1$.
Hence, it can be written as:-
Coefficient of ${x^3}$ in ${(1 - 2x)^{18}} = \dfrac{{18 \times 17 \times 16 \times 15!}}{{3!\left( {15} \right)!}} \times {1^{15}} \times {( - 2)^3}$
Simplifying it to get:- Coefficient of ${x^3}$ in ${(1 - 2x)^{18}} = \dfrac{{18 \times 17 \times 16}}{{3 \times 2}} \times {1^{15}} \times - 8$.
Simplifying it further by doing the calculation, we will get:-
Coefficient of ${x^3}$ in ${(1 - 2x)^{18}} = 3 \times 17 \times 16 \times - 8 = - 6528$. ………….(2)
Now, let us find similarly coefficient of ${x^2}$ in $a{(1 - 2x)^{18}}$ which will be:- $\dfrac{{18 \times 17 \times 16!}}{{2!\left( {16} \right)!}} \times {1^{16}} \times {( - 2)^2} \times a$.
So, coefficient of ${x^2}$ in $a{(1 - 2x)^{18}} = 9 \times 17 \times 4 \times a = 612a$. …………(3)
Now, let us find similarly coefficient of $x$ in $b{(1 - 2x)^{18}}$ which will be:- $\dfrac{{18 \times 17!}}{{1!\left( {17} \right)!}} \times {1^{17}} \times {( - 2)^1} \times b$.
So, coefficient of $x$ in $b{(1 - 2x)^{18}} = 18 \times - 2 \times b = - 36b$. …………(4)
Clubbing (2), (3) and (4), we will have:-
Coefficient of ${x^3}$ $(1 + ax + b{x^2}){(1 - 2x)^{18}} = - 6528 + 612a - 36b$.
Since, it is equal to 0, so, $- 6528 + 612a - 36b = 0$
This is equivalent to writing: $306a - 18b = 3264$ ……….(5)
Applying the same concept to find coefficient of ${x^4}$ in ${(1 - 2x)^{18}} + ax{(1 - 2x)^{18}} + b{x^2}{(1 - 2x)^{18}}$ is equivalent to finding the coefficient of ${x^4}$ in ${(1 - 2x)^{18}}$, ${x^3}$ in $a{(1 - 2x)^{18}}$ and ${x^2}$ in $b{(1 - 2x)^{18}}$.
Applying the formula as mentioned above in equation (1), we will get:-
Coefficient of ${x^4}$ in ${(1 - 2x)^{18}} = \dfrac{{18 \times 17 \times 16 \times 15 \times 14!}}{{4!\left( {14} \right)!}} \times {1^{14}} \times {( - 2)^4}$
Simplifying it to get:- Coefficient of ${x^4}$ in ${(1 - 2x)^{18}} = \dfrac{{18 \times 17 \times 16 \times 15}}{{4 \times 3 \times 2}} \times 16$.
Simplifying it further by doing the calculation, we will get:-
Coefficient of ${x^4}$ in ${(1 - 2x)^{18}} = 3 \times 17 \times 4 \times 15 \times 16 = 48960$. ………….(6)
Now, let us find similarly coefficient of ${x^3}$ in $a{(1 - 2x)^{18}}$ which will be:- $\dfrac{{18 \times 17 \times 16 \times 15!}}{{3!\left( {15} \right)!}} \times {1^{15}} \times {( - 2)^3} \times a$.
So, coefficient of ${x^3}$ in $a{(1 - 2x)^{18}} = 3 \times 17 \times 16 \times ( - 8) \times a = - 6528a$. …………(7)
Now, let us find similarly coefficient of ${x^2}$ in $b{(1 - 2x)^{18}}$ which will be:- $\dfrac{{18 \times 17 \times 16!}}{{2!\left( {16} \right)!}} \times {1^{16}} \times {( - 2)^2} \times b$.
So, coefficient of ${x^2}$ in $b{(1 - 2x)^{18}} = 9 \times 17 \times 4 \times b = 612b$. …………(8)
Clubbing (6), (7) and (8), we will have:-
Coefficient of ${x^4}$ $(1 + ax + b{x^2}){(1 - 2x)^{18}} = 48960 - 6528a + 612b$.
Since, it is equal to 0, so, $48960 - 6528a + 612b = 0$
This is equivalent to writing: $- 192a + 18b = - 1440$ ……….(9)
Adding (5) and (9) will result in:- $114a = 1824$
Hence, $a = 16$
Putting this in (5), we will get:-
$306 \times 16 - 18b = 3264$
$\Rightarrow - 18b = - 1632$
$\Rightarrow b = \dfrac{{1632}}{{18}} = \dfrac{{272}}{3}$.

So, the correct answer is “Option D”.

Additional Information: The formula we used in finding the coefficient, it one part which is $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ can be written as $^n{C_r}$ as well which basically means way of choosing r things from n things.

Note: The students must note that if they approach the problem directly before breaking it into three different parts, it may create a lot of confusion and hassle.