Question

$S{{=}^{404}}{{C}_{4}}{{-}^{4}}{{C}_{1}}{{.}^{303}}{{C}_{4}}{{+}^{4}}{{C}_{2}}{{.}^{202}}{{C}_{4}}{{-}^{2}}{{C}_{3}}{{.}^{101}}{{C}_{4}}={{\left( 101 \right)}^{k}}$ , then k is equal to:
(a) 1
(b) 2
(c) 4
(d) 6

Answer 1

Notice the pattern of the expression given in the question and relate the expression with the coefficient of ${{x}^{4}}$ in the expansion of $^{4}{{C}_{0}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{4}}{{-}^{4}}{{C}_{1}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{3}}{{+}^{4}}{{C}_{2}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{2}}{{-}^{4}}{{C}_{3}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{1}}$ . If we add and subtract one from this, we find that it is the expansion of ${{\left( 1-{{\left( 1+x \right)}^{101}} \right)}^{4}}-1$ . So, basically, we have to find the coefficient of ${{x}^{4}}$ in the expansion of ${{\left( 1-{{\left( 1+x \right)}^{101}} \right)}^{4}}-1$ .

Complete step-by-step answer:
Let us start with the solution to the above question. We can write the expression given in the question as:

& \text{co-efficient of }{{\text{x}}^{4}}\text{ in the expansion } \\\ & ^{4}{{C}_{0}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{4}}{{-}^{4}}{{C}_{1}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{3}}{{+}^{4}}{{C}_{2}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{2}}{{-}^{4}}{{C}_{3}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{1}} \\\ \end{aligned}$$ Now we will solve the expansion part and finally move on finding the required coefficient. So, let us first solve $$^{4}{{C}_{0}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{4}}{{-}^{4}}{{C}_{1}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{3}}{{+}^{4}}{{C}_{2}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{2}}{{-}^{4}}{{C}_{3}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{1}}$$ We will add and subtract 1 from the expression, on doing so, we get $$^{4}{{C}_{0}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{4}}{{-}^{4}}{{C}_{1}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{3}}{{+}^{4}}{{C}_{2}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{2}}{{-}^{4}}{{C}_{3}}.{{\left( {{\left( 1+x \right)}^{101}} \right)}^{1}}+1-1$$ 1 can be written as: $^{4}{{C}_{0}}{{\left( {{\left( 1+x \right)}^{101}} \right)}^{0}}$ . $$\begin{gathered} ^4{C_0}{\left( {{{\left( {1 + x} \right)}^{101}}} \right)^4}{ - ^4}{C_1}{\left( {{{\left( {1 + x} \right)}^{101}}} \right)^3}{ + ^4}{C_2}{\left( {{{\left( {1 + x} \right)}^{101}}} \right)^2} \\\ { - ^4}{C_3}{\left( {{{\left( {1 + x} \right)}^{101}}} \right)^1}{ + ^4}{C_0}{\left( {{{\left( {1 + x} \right)}^{101}}} \right)^0} - 1 \\\ \end{gathered} $$ As we can see, the above expression is the expansion of ${{\left( 1-{{\left( 1+x \right)}^{101}} \right)}^{4}}-1$ . We know that $${{\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}}}$$ . So, if we use this in our in our expression we get: ${{\left( 1-{{\left( 1+x \right)}^{101}} \right)}^{4}}-1$ $={{\left( 1{{-}^{101}}{{C}_{0}}{{x}^{0}}{{-}^{101}}{{C}_{1}}x-........ \right)}^{4}}-1$ . Now we will take x common and take it out of the bracket of forth power and considering the negative sign, we get $={{x}^{4}}{{\left( 1{{-}^{101}}{{C}_{0}}\times \dfrac{1}{x}{{-}^{101}}{{C}_{1}}\times 1-........ \right)}^{4}}-1$ So, the expression given in the question can be moulded as the coefficient of ${{x}^{4}}$ in the expansion of ${{x}^{4}}{{\left( 1{{-}^{101}}{{C}_{0}}\times \dfrac{1}{x}{{-}^{101}}{{C}_{1}}\times 1-........ \right)}^{4}}-1$ which can further be written as constant term in the expansion of ${{\left( 1{{-}^{101}}{{C}_{0}}\times \dfrac{1}{x}{{-}^{101}}{{C}_{1}}\times 1-........ \right)}^{4}}$ , as ${{x}^{4}}$ is getting multiplied with it and the separate -1 term can never have ${{x}^{4}}$ . So, the constant term is: ${{\left( ^{101}{{C}_{1}} \right)}^{4}}={{\left( \dfrac{101!}{\left( 101-1 \right)!1!} \right)}^{4}}={{101}^{4}}$ So, comparing it with the equation given in the question, we get ${{101}^{4}}={{\left( 101 \right)}^{k}}$ Therefore, k=4. Hence, the answer to the above question is option (c). **So, the correct answer is “Option c”.** **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.