
If \[{\left( {1 + ax} \right)^n} = 1 + 6x + \left( {\dfrac{{27}}{2}} \right){x^2} + {a^n}{x^n}\], then the values of a and n are respectively
A. 2 , 3
B. 3,2
C. $\dfrac{3}{2}$ , 4
D. 1 , 6
E. $\dfrac{3}{2}$ , 6
Answer
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Hint: In this equation, we have to find out the binomial coefficient of the above equation. To find out we need to apply the binomial formula for \[{\left( {1 + ax} \right)^n}\] after applying the formula also compare the above-given equation to find the respective variable of coefficients.
Formula Used:
The formula for binomial expansion of \[{\left( {1 + ax} \right)^n}\]is:
\[{\left( {1 + ax} \right)^n} = n\left( {ax} \right) + \dfrac{{n.\left( {n - 1} \right)}}{{2!}}{\left( {ax} \right)^2} + ..... + \dfrac{{n\left( {n - 1} \right)..(n - r + 1)}}{{r!}}{\left( {ax} \right)^r}\]
Complete step by step Solution:
Given \[{\left( {1 + ax} \right)^n} = 1 + 6x + \left( {\dfrac{{27}}{2}} \right){x^2} + ........... + {a^n}{x^n}\]
The formula for binomial expansion of \[{\left( {1 + ax} \right)^n}\]is:
\[{\left( {1 + ax} \right)^n} = n\left( {ax} \right) + \dfrac{{n.\left( {n - 1} \right)}}{{2!}}{\left( {ax} \right)^2} + ..... + \dfrac{{n\left( {n - 1} \right)..(n - r + 1)}}{{r!}}{\left( {ax} \right)^r}\]
Taking the above two equation and simplifying them
\[{\left( {1 + ax} \right)^n} = 1 + 6x + \left( {\dfrac{{27}}{2}} \right){x^2} + ........ + {a^n}{x^n}\]
\[{\left( {1 + ax} \right)^n} = {}^n{c_0} + {}^n{c_1}ax + {}^n{c_2}{\left( {ax} \right)^2} + ..... + n{c_n}{a^n}{x^n}\]
On comparing the coefficient of like powers in Equation (1) and (2)
\[n{c_1}ax = 6x;{}^n{c_2}{\left( {ax} \right)^2} = \dfrac{{27}}{2}{x^2};.....\]
\[\; \Rightarrow an = 6\]
This can also be written as \[n = \dfrac{6}{a}\]
\[ \Rightarrow n(n - 1){a^2} = 27\]
\[{n^2}{a^2} - n{a^2} = 27\]
In this above equation (4) putting the value of n from equation (3)
\[{\left( {\dfrac{6}{a}} \right)^2}{a^2} - \left( {\dfrac{6}{a}} \right){a^2} = 27\]
Further solving this equation
\[36 - 6a = 27\]
\[36 - 27 = 6a\]
\[9 = 6a\]
Further simplifying this equation we get
$ \Rightarrow a = \dfrac{9}{6} = \dfrac{3}{2}$
Now substituting the value of a in equation (3) to find value of n
\[\; \Rightarrow an = 6\]
\[\; \Rightarrow \dfrac{3}{2}n = 6\]
$n = \dfrac{{6 \times 2}}{3} = 4$
Hence, the correct option is C.
Note: To solve the above equation we can simply remember the binomial expression and also how to compare the equation with the given and find the coefficient. Rest in the question there is only simplification needed and finding the values.
Formula Used:
The formula for binomial expansion of \[{\left( {1 + ax} \right)^n}\]is:
\[{\left( {1 + ax} \right)^n} = n\left( {ax} \right) + \dfrac{{n.\left( {n - 1} \right)}}{{2!}}{\left( {ax} \right)^2} + ..... + \dfrac{{n\left( {n - 1} \right)..(n - r + 1)}}{{r!}}{\left( {ax} \right)^r}\]
Complete step by step Solution:
Given \[{\left( {1 + ax} \right)^n} = 1 + 6x + \left( {\dfrac{{27}}{2}} \right){x^2} + ........... + {a^n}{x^n}\]
The formula for binomial expansion of \[{\left( {1 + ax} \right)^n}\]is:
\[{\left( {1 + ax} \right)^n} = n\left( {ax} \right) + \dfrac{{n.\left( {n - 1} \right)}}{{2!}}{\left( {ax} \right)^2} + ..... + \dfrac{{n\left( {n - 1} \right)..(n - r + 1)}}{{r!}}{\left( {ax} \right)^r}\]
Taking the above two equation and simplifying them
\[{\left( {1 + ax} \right)^n} = 1 + 6x + \left( {\dfrac{{27}}{2}} \right){x^2} + ........ + {a^n}{x^n}\]
\[{\left( {1 + ax} \right)^n} = {}^n{c_0} + {}^n{c_1}ax + {}^n{c_2}{\left( {ax} \right)^2} + ..... + n{c_n}{a^n}{x^n}\]
On comparing the coefficient of like powers in Equation (1) and (2)
\[n{c_1}ax = 6x;{}^n{c_2}{\left( {ax} \right)^2} = \dfrac{{27}}{2}{x^2};.....\]
\[\; \Rightarrow an = 6\]
This can also be written as \[n = \dfrac{6}{a}\]
\[ \Rightarrow n(n - 1){a^2} = 27\]
\[{n^2}{a^2} - n{a^2} = 27\]
In this above equation (4) putting the value of n from equation (3)
\[{\left( {\dfrac{6}{a}} \right)^2}{a^2} - \left( {\dfrac{6}{a}} \right){a^2} = 27\]
Further solving this equation
\[36 - 6a = 27\]
\[36 - 27 = 6a\]
\[9 = 6a\]
Further simplifying this equation we get
$ \Rightarrow a = \dfrac{9}{6} = \dfrac{3}{2}$
Now substituting the value of a in equation (3) to find value of n
\[\; \Rightarrow an = 6\]
\[\; \Rightarrow \dfrac{3}{2}n = 6\]
$n = \dfrac{{6 \times 2}}{3} = 4$
Hence, the correct option is C.
Note: To solve the above equation we can simply remember the binomial expression and also how to compare the equation with the given and find the coefficient. Rest in the question there is only simplification needed and finding the values.
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