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

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.