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Question: If rth term in the expansion of \( {\left( {{x^2} + \dfrac{1}{x}} \right)^{12}} \) is independent of...

If rth term in the expansion of (x2+1x)12{\left( {{x^2} + \dfrac{1}{x}} \right)^{12}} is independent of x, then r =
a. 9
b. 8
c. 10
d. none of these

Explanation

Solution

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=nC0anb0+nC1an−1b1+nC2an−2b2+...+nCna0bn{(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 (x2+1x)12{\left( {{x^2} + \dfrac{1}{x}} \right)^{12}}
Here we have n = 12 a=x2a = {x^2} and b=1xb = \dfrac{1}{x} . Substituting all the values in the formula (a+b)n=nC0anb0+nC1an−1b1+nC2an−2b2+...+nCna0bn{(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
(x2+1x)12=12C0(x2)12(1x)0+12C1(x2)12−1(1x)1+12C2(x2)12−2(1x)2 \+12C3(x2)12−3(1x)3+12C4(x2)12−4(1x)4+12C5(x2)12−5(1x)5 \+12C6(x2)12−6(1x)6+12C7(x2)12−7(1x)7+12C8(x2)12−8(1x)8 \+12C9(x2)12−9(1x)9+12C10(x2)12−10(1x)10+12C11(x2)12−11(1x)11+12C12(x2)0(1x)12  {\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 nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} and we use this formula to simplify the terms and so we have
⇒(x2+1x)12=x24+12!11!1!x22(1x)+12!10!2!x20(1x2) \+12!9!3!x18(1x3)+12!8!4!x16(1x4)+12!7!5!x14(1x5)+12!6!6!x12(1x6) \+12!5!7!x10(1x7)+12!4!8!x8(1x8)+12!3!9!x6(1x9)+12!2!10!x4(1x10) \+12!1!11!x2(1x11)+(1x12)  \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×(n−1)×(n−2)×...×2×1n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 by using this formula we calculate the factorial terms and we have
⇒(x2+1x)12=x24+12x21+66x18+220x15+495x12+792x9+924x6 \+792x3+495+220x−3+66x−6+12x−9+x−12   \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=nC0anb0+nC1an−1b1+nC2an−2b2+...+nCna0bn{(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.