Question

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

Answer 1

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