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)!}} $