Question

If rth term in the expansion of $ {\left( {{x^2} + \dfrac{1}{x}} \right)^{12}} $ is independent of x, then r =
a. 9
b. 8
c. 10
d. none of these

Answer 1

Hint: To find the value of the r th term we use the formula binomial expansion and the formula is given as $ {(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n} $ . We apply the binomial expansion where n is 12. The r th term is independent of x.

Complete step-by-step answer:
Here we have to find the rth term and they have mentioned that the rth term is independent of x where it does not contain any x term, we can say it as a constant term. To solve this question, we use the formula of binomial expansion and after that we use a factorial formula to solve further.
Now we apply binomial expansion to $ {\left( {{x^2} + \dfrac{1}{x}} \right)^{12}} $
Here we have n = 12 $ a = {x^2} $ and $ b = \dfrac{1}{x} $ . Substituting all the values in the formula $ {(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n} $
So we have
 $
  {\left( {{x^2} + \dfrac{1}{x}} \right)^{12}} = {}^{12}{C_0}{\left( {{x^2}} \right)^{12}}{\left( {\dfrac{1}{x}} \right)^0} + {}^{12}{C_1}{\left( {{x^2}} \right)^{12 - 1}}{\left( {\dfrac{1}{x}} \right)^1} + {}^{12}{C_2}{\left( {{x^2}} \right)^{12 - 2}}{\left( {\dfrac{1}{x}} \right)^2} \\
   + {}^{12}{C_3}{\left( {{x^2}} \right)^{12 - 3}}{\left( {\dfrac{1}{x}} \right)^3} + {}^{12}{C_4}{\left( {{x^2}} \right)^{12 - 4}}{\left( {\dfrac{1}{x}} \right)^4} + {}^{12}{C_5}{\left( {{x^2}} \right)^{12 - 5}}{\left( {\dfrac{1}{x}} \right)^5} \\
   + {}^{12}{C_6}{\left( {{x^2}} \right)^{12 - 6}}{\left( {\dfrac{1}{x}} \right)^6} + {}^{12}{C_7}{\left( {{x^2}} \right)^{12 - 7}}{\left( {\dfrac{1}{x}} \right)^7} + {}^{12}{C_8}{\left( {{x^2}} \right)^{12 - 8}}{\left( {\dfrac{1}{x}} \right)^8} \\
   + {}^{12}{C_9}{\left( {{x^2}} \right)^{12 - 9}}{\left( {\dfrac{1}{x}} \right)^9} + {}^{12}{C_{10}}{\left( {{x^2}} \right)^{12 - 10}}{\left( {\dfrac{1}{x}} \right)^{10}} + {}^{12}{C_{11}}{\left( {{x^2}} \right)^{12 - 11}}{\left( {\dfrac{1}{x}} \right)^{11}} + {}^{12}{C_{12}}{\left( {{x^2}} \right)^0}{\left( {\dfrac{1}{x}} \right)^{12}} \\
  $
We know the formula $ {}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} $ and we use this formula to simplify the terms and so we have
 $
   \Rightarrow {\left( {{x^2} + \dfrac{1}{x}} \right)^{12}} = {x^{24}} + \dfrac{{12!}}{{11!1!}}{x^{22}}\left( {\dfrac{1}{x}} \right) + \dfrac{{12!}}{{10!2!}}{x^{20}}\left( {\dfrac{1}{{{x^2}}}} \right) \\
   + \dfrac{{12!}}{{9!3!}}{x^{18}}\left( {\dfrac{1}{{{x^3}}}} \right) + \dfrac{{12!}}{{8!4!}}{x^{16}}\left( {\dfrac{1}{{{x^4}}}} \right) + \dfrac{{12!}}{{7!5!}}{x^{14}}\left( {\dfrac{1}{{{x^5}}}} \right) + \dfrac{{12!}}{{6!6!}}{x^{12}}\left( {\dfrac{1}{{{x^6}}}} \right) \\
   + \dfrac{{12!}}{{5!7!}}{x^{10}}\left( {\dfrac{1}{{{x^7}}}} \right) + \dfrac{{12!}}{{4!8!}}{x^8}\left( {\dfrac{1}{{{x^8}}}} \right) + \dfrac{{12!}}{{3!9!}}{x^6}\left( {\dfrac{1}{{{x^9}}}} \right) + \dfrac{{12!}}{{2!10!}}{x^4}\left( {\dfrac{1}{{{x^{10}}}}} \right) \\
   + \dfrac{{12!}}{{1!11!}}{x^2}\left( {\dfrac{1}{{{x^{11}}}}} \right) + \left( {\dfrac{1}{{{x^{12}}}}} \right) \\
  $
By simplifying we have
For the simplification we need n factorial formula since we factorial therefore the formula is $ n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 $ by using this formula we calculate the factorial terms and we have
 $
   \Rightarrow {\left( {{x^2} + \dfrac{1}{x}} \right)^{12}} = {x^{24}} + 12{x^{21}} + 66{x^{18}} + 220{x^{15}} + 495{x^{12}} + 792{x^9} + 924{x^6} \\
   + 792{x^3} + 495 + 220{x^{ - 3}} + 66{x^{ - 6}} + 12{x^{ - 9}} + {x^{ - 12}} \;
  $
On the simplification we have a polynomial, we can see the only one term which is independent of x and we can say it has a constant. The constant term is the 9 th term.
Therefore, the r th term is 9 th term
So, the correct answer is “Option A”.

Note: To solve this type of this we use the binomial expansion formula and the formula is defined as $ {(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n} $ by substituting the value of a, b and n we can calculate the solution for this question. We will determine the r th term on the basis of what they have given.