Question

The first 3 terms in the expansion of ${\left( {1 + ax} \right)^n}\left( {n \ne 0} \right)$ are 1, $6x$ and $16{x^2}$ . Then the values of a and n are respectively
A. 2 and 9
B. 3 and 2
C. 2/3 and 9
D. 3/2 and 6

Answer 1

Hint : First expand ${\left( {1 + ax} \right)^n}$ following the binomial expansion which is mentioned below. And take out the first 3 terms of the expansion and compare them with the given values 1, $6x$ and $16{x^2}$ respectively. From this we get the values of a and n.
Formula used:
Binomial expansion of ${\left( {x + y} \right)^n}$ is \mathop \sum \limits_{r = 0}^n {}_{}^nC_r^{}.{x^{n - r}}.{y^r} , where r must always be less than or equal to n.

Complete step-by-step answer :
We are given that the first 3 terms in the expansion of ${\left( {1 + ax} \right)^n}\left( {n \ne 0} \right)$ are 1, $6x$ and $16{x^2}$ .
We have to find the value of ‘a’ and ‘n’.
First we are expanding ${\left( {1 + ax} \right)^n}$ .
Binomial expansion of ${\left( {1 + ax} \right)^n}$ is \mathop \sum \limits_{r = 0}^n {}_{}^nC_r^{}.{\left( 1 \right)^{n - r}}\left( {ax} \right)
$\Rightarrow {}_{}^nC_0^{}{\left( 1 \right)^{n - 0}}{\left( {ax} \right)^0} + {}_{}^nC_1^{}{\left( 1 \right)^{n - 1}}{\left( {ax} \right)^1} + {}_{}^nC_2^{}{\left( 1 \right)^{n - 2}}{\left( {ax} \right)^2} + ....$
The value of ${}_{}^nC_0^{}$ is 1, value of ${}_{}^nC_1^{}$ is n and the value of ${}_{}^nC_2^{}$ is $\dfrac{{n\left( {n - 1} \right)}}{2}$
Substituting the above values, we get
$\Rightarrow 1 \times {\left( 1 \right)^n} + n \times {\left( 1 \right)^{n - 1}}\left( {ax} \right) + \left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right){\left( 1 \right)^{n - 2}}{\left( {ax} \right)^2} + ....$
1 raised to the power of anything will result in 1 itself.
$\Rightarrow \left( {1 \times 1} \right) + \left( {n \times 1 \times ax} \right) + \left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right) \times 1 \times {\left( {ax} \right)^2} + ....$
$\Rightarrow 1 + nax + \left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right){\left( {ax} \right)^2} + ....$
As we can see the first term is 1, the second term is $nax$ and the third term is $\left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right){\left( {ax} \right)^2}$ of the expansion. But we are given that 1, $6x$ and $16{x^2}$ are the first three terms.
This means $nax = 6x,\left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right){\left( {ax} \right)^2} = 16{x^2}$
$nax = 6x$
$\Rightarrow na = 6$
$\Rightarrow a = \dfrac{6}{n} \Rightarrow eq\left( 1 \right)$
$\left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right){\left( {ax} \right)^2} = 16{x^2}$
Substituting the value of ‘a’ from equation 1
$\Rightarrow \left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right){\left( {\dfrac{6}{n} \times x} \right)^2} = 16{x^2}$
$\Rightarrow \left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right)\dfrac{{36{x^2}}}{{{n^2}}} = 16{x^2}$
Cancelling ${x^2}$ on either side
$\Rightarrow \left( {\dfrac{{n\left( {n - 1} \right)}}{2}} \right)\dfrac{{36}}{{{n^2}}} = 16$
$\Rightarrow \dfrac{{n\left( {n - 1} \right)}}{2} = 16 \times \dfrac{{{n^2}}}{{36}}$
$\Rightarrow \dfrac{{{n^2} - n}}{2} = \dfrac{{4{n^2}}}{9}$
On cross multiplication, we get
$\Rightarrow 9\left( {{n^2} - n} \right) = 2 \times 4{n^2}$
$\Rightarrow 9{n^2} - 9n = 8{n^2}$
$\Rightarrow 9{n^2} - 9n - 8{n^2} = 0$
$\Rightarrow {n^2} - 9n = 0$
$\Rightarrow n\left( {n - 9} \right) = 0$
N cannot be zero, so $\left( {n - 9} \right) = 0$
$\therefore n = 9$
The value of n is 9 and the value of a is $\dfrac{6}{n} = \dfrac{6}{9} = \dfrac{2}{3}$
Hence, the correct option is Option C, a is $\dfrac{2}{3}$ and n is 9.
So, the correct answer is “Option C”.

Note : Instead of expanding ${\left( {1 + ax} \right)^n}$ , we can directly write the 1st, 2nd and 3rd terms using the general formula of the terms of a binomial expression which is ${T_{r + 1}} = {}_{}^nC_r^{}.{\left( 1 \right)^{n - r}}{\left( {ax} \right)^r}$ , where r starts from 0 and always less than or equal to n. The value of ${}_{}^nC_r^{}$ is $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$