Question

The coefficient of ${{t}^{4}}$ in the expansion of ${{\left( \dfrac{1-{{t}^{6}}}{1-t} \right)}^{3}}$ is:
(a) 12
(b) 15
(c) 10
(d) 14

Answer 1

Hint: We can represent ${{\left( \dfrac{1-{{t}^{6}}}{1-t} \right)}^{3}}$ as ${{\left( 1-{{t}^{6}} \right)}^{3}}{{\left( 1-t \right)}^{-3}}$ . If you expand ${{\left( 1-{{t}^{6}} \right)}^{3}}$ you will find that it contains terms with the power of t as multiple of 6, so the only useful term out of this is the constant term rest all will have power of t greater than 4 and as we know that the expansion of ${{\left( 1-t \right)}^{-3}}$ would contain all the powers of t but only the useful one will be the term with power of t as 4 as we have a constant term from the first part that will be multiplied with it. Find the terms and multiply them to get the answer.

Complete step-by-step answer:
Let us start with the solution to the above question. The binomial expansion of ${{\left( 1+x \right)}^{n}}$ is:
${{\left( 1+x \right)}^{n}}={{\text{ }}^{n}}{{\text{C}}_{0}}{{x}^{0}}{{+}^{n}}{{\text{C}}_{1}}{{x}^{1}}{{+}^{n}}{{\text{C}}_{2}}{{x}^{2}}+........{{+}^{n}}{{\text{C}}_{n}}{{x}^{n}}=\sum{^{n}{{C}_{r}}{{x}^{r}}}$
Using the above expansion if we substitute n by 3 and x by ${{t}^{6}}$ , we get
${{\left( 1-{{t}^{6}} \right)}^{3}}={{\text{ }}^{3}}{{\text{C}}_{0}}{{-}^{3}}{{\text{C}}_{1}}{{t}^{6}}{{+}^{3}}{{\text{C}}_{2}}{{t}^{12}}{{-}^{3}}{{\text{C}}_{3}}{{t}^{18}}$
Now let us move to the expansion of ${{\left( 1-t \right)}^{-3}}$ . We know that the expansion of ${{\left( 1+x \right)}^{n}}$ , where n is a negative integer is given by:
${{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}{4!}{{x}^{4}}...................$
So, if we substitute n by -3 and x by –t, we get
${{\left( 1-t \right)}^{-3}}=1+3t+\dfrac{\left( -3 \right)\left( -3-1 \right)}{2!}{{t}^{2}}-\dfrac{\left( -3 \right)\left( -3-1 \right)\left( -3-2 \right)}{3!}{{t}^{3}}+\dfrac{\left( -3 \right)\left( -3-1 \right)\left( -3-2 \right)\left( -3-3 \right)}{4!}{{t}^{4}}......$
$\Rightarrow {{\left( 1-t \right)}^{-3}}=1+3t+\dfrac{12}{2!}{{t}^{2}}+\dfrac{60}{3!}{{t}^{3}}+\dfrac{360}{4!}{{t}^{4}}...................$
Now let us move to the expression given in the question. The expression given is ${{\left( \dfrac{1-{{t}^{6}}}{1-t} \right)}^{3}}$ and this can be represented as ${{\left( 1-{{t}^{6}} \right)}^{3}}{{\left( 1-t \right)}^{-3}}$ . If we look at its expansion using the above results, we get
${{\left( 1-{{t}^{6}} \right)}^{3}}{{\left( 1-t \right)}^{-3}}=\left( ^{3}{{\text{C}}_{0}}{{-}^{3}}{{\text{C}}_{1}}{{t}^{6}}{{+}^{3}}{{\text{C}}_{2}}{{t}^{12}}{{-}^{3}}{{\text{C}}_{3}}{{t}^{18}} \right)\left( 1+3t+\dfrac{12}{2!}{{t}^{2}}+\dfrac{60}{3!}{{t}^{3}}+\dfrac{360}{4!}{{t}^{4}}....... \right)$
If we see that only the term from the first bracket of our use is the constant term, rest all have power of three greater than 4. If we take the constant term, the term of our use from the second bracket is the term with power of t=4, so that we get a term with power of t=4. So, the term with power of t as 4 is:
$^{3}{{C}_{0}}\times \dfrac{360}{4!}{{t}^{4}}=1\times \dfrac{360}{24}{{t}^{4}}=15{{t}^{4}}$
Therefore, the coefficient of ${{t}^{4}}$ is 15. Hence, the answer to the above question is option (b).

Note: Always be careful with the signs that appear in the expansions, as the students are generally finding signs to be a concern while using the binomial expansions. Also, be careful about the calculation part, as in general cases, the questions involving binomial expansion contain very long and complex calculations due to the presence of factorial terms. You should also know that the binomial coefficient and actual coefficients might or might not be the same. For example: in the expansion of ${{\left( 1+3x \right)}^{3}}$ , the binomial coefficient of ${{x}^{3}}$ is $^{3}{{C}_{3}}=1$ and coefficient is 27.