Question

In the binomial expansion of \[{\left( {1 + ax} \right)^n}\] , where \[a\] and \[n\] are constants, the coefficient of \[x\] is \[15\] .The coefficient of \[{x^2}\] and of \[{x^3}\] are equal. What is the value of \[a\] and of \[n\] ?

Answer 1

Hint: In order to find the value of \[a\] and \[n\] we will first use the formula of binomial expansion i.e., \[{\left( {1 + ax} \right)^n} = 1 + n\left( {ax} \right) + \dfrac{{n \cdot \left( {n - 1} \right)}}{{2!}}{\left( {ax} \right)^2} + ..... + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)...\left( {n - r + 1} \right)}}{{r!}}{\left( {ax} \right)^r} + .....\] Now first we will equate the coefficient of \[x\] with \[15\] . After that we will equate the coefficients of \[{x^2}\] and of \[{x^3}\] and simplify to get the value of \[a\] and of \[n\] and hence we get the required result.

Complete step by step answer:
We know that
The formula for the binomial expansion of \[{\left( {1 + ax} \right)^n}\] is:
\[{\left( {1 + ax} \right)^n} = 1 + n\left( {ax} \right) + \dfrac{{n \cdot \left( {n - 1} \right)}}{{2!}}{\left( {ax} \right)^2} + \dfrac{{n \cdot \left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{\left( {ax} \right)^3} + ..... + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)...\left( {n - r + 1} \right)}}{{r!}}{\left( {ax} \right)^r} + .....\]
Therefore, the coefficient of \[x\] is \[an\]
the coefficient of \[{x^2}\] is \[\dfrac{{n\left( {n - 1} \right)}}{{2!}}{\left( a \right)^2}\]
and the coefficient of \[{x^3}\] is \[\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{\left( a \right)^3}\]
Now it is given that
the coefficient of \[x\] is \[15\]
\[ \Rightarrow an = 15{\text{ }} - - - \left( i \right)\]
And the coefficient of \[{x^2}\] and of \[{x^3}\] are equal
\[ \Rightarrow \dfrac{{n\left( {n - 1} \right)}}{{2!}}{\left( a \right)^2} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{\left( a \right)^3}{\text{ }} - - - \left( {ii} \right)\]
Now, from equation \[\left( {ii} \right)\] on cancelling the like terms, we get
\[\dfrac{1}{{2!}} = \dfrac{{n - 2}}{{3!}}a\]
We know that
\[n! = n \cdot \left( {n - 1} \right)!\]
Therefore, we get
\[\dfrac{1}{{2!}} = \dfrac{{n - 2}}{{3 \times 2!}}a\]
\[ \Rightarrow 1 = \dfrac{{n - 2}}{3}a\]
On multiplying by \[3\] on both sides, we get
\[3 = \left( {n - 2} \right)a\]
\[ \Rightarrow 3 = an - 2a\]
Now on substituting the value of \[an\] from equation \[\left( i \right)\] we get
\[ \Rightarrow 3 = 15 - 2a\]
\[ \Rightarrow 2a = 12\]
On dividing by \[2\] we get
\[ \Rightarrow a = 6\]
So, from equation \[\left( i \right)\] we have
\[an = 15\]
\[ \Rightarrow n = \dfrac{{15}}{a}\]
On substituting the value of \[a\] we get
\[ \Rightarrow n = \dfrac{{15}}{6}\]
\[ \Rightarrow n = 2.5\]
Hence, we get the value of \[a\] as \[6\] and the value of \[n\] as \[2.5\]

Note:
Students must know the binomial expansion of \[{\left( {1 + ax} \right)^n}\] .Also note that the formula for binomial expansion of \[{\left( {1 + ax} \right)^n}\] can also be written as:
\[{\left( {1 + ax} \right)^n} = {}^n{C_0}\left( 1 \right){\left( {ax} \right)^0} + {}^n{C_1}{\left( 1 \right)^{n - 1}}{\left( {ax} \right)^1} + {}^n{C_2}{\left( 1 \right)^{n - 2}}{\left( {ax} \right)^2} + ... + {}^n{C_n}{\left( 1 \right)^0}{\left( {ax} \right)^n}\]
where \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
Here \[{}^n{C_0},{}^n{C_1},{}^n{C_2},....,{}^n{C_n}\] are called binomial coefficients. And the total number of terms in the expansion of \[{\left( {1 + ax} \right)^n}\] are \[\left( {n + 1} \right)\] .These are a few points about binomial expansion. You should keep all these things in your mind while solving the questions of binomial expansion.