Question

Find the 13th term in the expansion of ${\left( {9x - \dfrac{1}{{3\sqrt x }}} \right)^{18}},x \ne 0$

Answer 1

Here we will use binomial term expansion and expand the given term to get its desired value.
The binomial expansion is given:-

$${\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^0}{\left( b \right)^n} + {}^n{C_1}{\left( a \right)^1}{\left( b \right)^{n - 1}} + {}^n{C_2}{\left( a \right)^2}{\left( b \right)^{n - 2}} + ..................... + {}^n{C_n}{\left( a \right)^n}{\left( b \right)^0} \\\ \\\$$

Complete step-by-step answer:
The given expansion is:
${\left( {9x - \dfrac{1}{{3\sqrt x }}} \right)^{18}}$
Now since we know that the general expansion of ${\left( {a + b} \right)^n}$ is given by:
${T_{r + 1}} = {}^n{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}$
Now since we have to find the 13th term of the expression
Therefore,

$$r = 12 \\\ n = 18 \\\ a = 9x \\\ b = - \dfrac{1}{{3\sqrt x }} \\\$$

Putting in the values we get:
${T_{12 + 1}} = {}^{18}{C_{12}}{\left( {9x} \right)^{18 - 12}}{\left( { - \dfrac{1}{{3\sqrt x }}} \right)^{12}}$
Now since we can write
${\left( { - \dfrac{1}{{3\sqrt x }}} \right)^{12}} = {\left( {\dfrac{{ - 1}}{3}} \right)^{12}}{\left[ {{{\left( {\dfrac{1}{x}} \right)}^{\dfrac{1}{2}}}} \right]^{12}}$
Therefore putting in the values we get:-
${T_{13}} = {}^{18}{C_{12}}{\left( {9x} \right)^6}{\left( {\dfrac{{ - 1}}{3}} \right)^{12}}{\left[ {{{\left( {\dfrac{1}{x}} \right)}^{\dfrac{1}{2}}}} \right]^{12}}$
On simplifying it we get:-
${T_{13}} = {}^{18}{C_{12}}{\left( {{3^2}} \right)^6}{\left( x \right)^6}{\left( {\dfrac{1}{3}} \right)^{12}}{\left( {\dfrac{1}{x}} \right)^6}$
Cancelling the terms we get:-
${T_{13}} = {}^{18}{C_{12}}$
Now we know that,
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Therefore,
${T_{13}} = \dfrac{{18!}}{{12! \times 6!}}$
Now expanding the factorials we get:-
${T_{13}} = \dfrac{{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12!}}{{12! \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}$
Cancelling the terms and solving it further we get:-
${T_{13}} = 17 \times 2 \times 3 \times 14 \times 13$
Multiplying the terms we get:-
${T_{13}} = 18564$

Hence 13th term is 18564

Note: The general expansion of ${\left( {a + b} \right)^n}$ is given by:
${T_{r + 1}} = {}^n{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}$In order to ease the calculation, one should use the above formula otherwise we can also solve this question by the following formula:

$${\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^0}{\left( b \right)^n} + {}^n{C_1}{\left( a \right)^1}{\left( b \right)^{n - 1}} + {}^n{C_2}{\left( a \right)^2}{\left( b \right)^{n - 2}} + ..................... + {}^n{C_n}{\left( a \right)^n}{\left( b \right)^0} \\\ \\\$$