Question

Find the coefficient of ${x^2}$ in ${\left( {\sqrt {\dfrac{x}{3}} + \dfrac{3}{{2x}}} \right)^{10}}$.

Answer 1

The question given above is of binomial expansion. To find the coefficient of such expressions, first we write the formula of the general term of binomial expansion. Then we try to separate the variable part with the coefficient part. Since we have to find the coefficient of ${x^2}$, we put the power part of $x$ equal to $2$. We get the value of the position of term whose variable is ${x^2}$. Put that value in the coefficient part to get the required coefficient.

Formula used: The formula for general term of a binomial expansion is Let ${T_{r + 1}}$ be the number of the term, $r$ be the position of term and $x$ be the variable, power of binomial is $n$. So ${(r + 1)^{th}}$ term of expansion ${\left( {a + b} \right)^n}$ is given by, ${T_{r + 1}} = \mathop C\limits_r^n {a^{n - r}}{b^r}$.

Complete step-by-step answer:
First we write the formula of general term of binomial expansion. Let ${T_{r + 1}}$ be the number of the term,
$r$ be the position of term and $x$ be the variable, power of binomial$n = 10$. So,
${T_{r + 1}} = \mathop C\limits_r^n {\sqrt {\dfrac{x}{3}} ^{n - r}}{\dfrac{3}{{2x}}^r}$
Now we try to separate $x$ from other.

$$\Rightarrow {T_{r + 1}} = \mathop C\limits_r^{10} {\left( {\dfrac{x}{3}} \right)^{\dfrac{{10 - r}}{2}}}\dfrac{{{3^r}}}{{{2^r}{x^r}}} \\\ \Rightarrow {T_{r + 1}} = \mathop C\limits_r^{10} \left( {\dfrac{{{x^{\dfrac{{10 - r}}{2}}}}}{{{3^{\dfrac{{10 - r}}{2}}}}}} \right)\dfrac{{{3^r}}}{{{2^r}{x^r}}} \\\ \Rightarrow {T_{r + 1}} = \mathop C\limits_r^{10} \left( {\dfrac{{{3^r}}}{{{3^{\dfrac{{10 - r}}{2}}}{2^r}}}} \right)\dfrac{{{x^{\dfrac{{10 - r}}{2}}}}}{{{x^r}}} \\\ \Rightarrow {T_{r + 1}} = \mathop C\limits_r^{10} \left( {\dfrac{{{3^r}}}{{{3^{\dfrac{{10 - r}}{2}}}{2^r}}}} \right){x^{\dfrac{{10 - r}}{{2r}}}} \\\$$

Now, since we have to find coefficient of ${x^2}$, we put the power part of $x$ equal to $2$, so,

$$\dfrac{{10 - r}}{{2r}} = 2 \\\ \Rightarrow 10 - r = 4r \\\ \Rightarrow 5r = 10 \\\ \Rightarrow r = 2 \\\$$

Hence, ${T_{r + 1}}$=${T_{2 + 1}} = {T_3}$. So ${3^{rd}}$ term have variable ${x^2}$.
Now, to get the coefficient of variable ${x^2}$we put the value of $r = 2$ in $\mathop C\limits_r^{10} \left( {\dfrac{{{3^r}}}{{{3^{\dfrac{{10 - r}}{2}}}{2^r}}}} \right){x^{\dfrac{{10 - r}}{{2r}}}}$.
So,

$$\Rightarrow \mathop C\limits_2^{10} \left( {\dfrac{{{3^2}}}{{{3^{\dfrac{{10 - 2}}{2}}}{2^2}}}} \right) \\\ \Rightarrow \mathop C\limits_2^{10} \left( {\dfrac{{{3^2}}}{{{3^{\dfrac{8}{2}}}{2^2}}}} \right) \\\ \Rightarrow \mathop C\limits_2^{10} \left( {\dfrac{{{3^2}}}{{{3^4}{2^2}}}} \right) \\\ \Rightarrow \mathop C\limits_2^{10} \left( {\dfrac{1}{{{3^2}{2^2}}}} \right) \\\$$

So, the coefficient of ${x^2}$ is $\mathop C\limits_2^{10} \left( {\dfrac{1}{{{3^2}{2^2}}}} \right)$.

Note: The Binomial theorem tells us how to expand expressions of the form (a+b)ⁿ, for example, (x+y)⁷. The larger the power is, the harder it is to expand expressions like this directly. But with the Binomial theorem, the process is relatively fast. But one must be careful in calculation as there are chances to miss any term.