4.0 General term

  Binomial Theorem

The term ${}^n{C_r}{a^{n - r}}{x^r}$ is the $(r+1)$th term from beginning in the expansion of ${(a + x)^n}$ which is also called the general term and denoted by $${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{x^r}$$

Question 2. If the binomial coefficients of the ${(2r+4)^{th}}$ term and the ${(r-2)^{th}}$ term in the expansion of ${(1 + x)^{18}}$ are equal, find the value of $r$.

Solution: The coefficient of ${(2r+4)^{th}}$ term in the expansion of ${(1 + x)^{18}}$ is ${}^{18}{C_{2r + 3}}$ and the coefficient of ${(r-2)^{th}}$ term in the expansion of ${(1 + x)^{18}}$ is $^{18}{C_{r - 3}}$. Therefore, $$^{18}{C_{2r + 3}}{ = ^{18}}{C_{r - 3}}$$

Using property $(1)$, we can say that either $$\begin{equation} \begin{aligned} 2r + 3 = r - 3 \\ \Rightarrow r = - 6 \\\end{aligned} \end{equation} $$

or $$\begin{equation} \begin{aligned} 2r + 3 + r - 3 = 18 \\ \Rightarrow r = 6 \\\end{aligned} \end{equation} $$

Since $r \in N$, $r=6$ is the only solution.

Question 3. Find the coefficient of term independent of $x$ in the expansion of $${\left[ {\sqrt {\frac{x}{3}} + \frac{{\sqrt 3 }}{{2{x^2}}}} \right]^{10}}$$

Solution: General term is

$$\begin{equation} \begin{aligned} {T_{r+1}} = {}^{10}{C_r}{\left( {\sqrt {\frac{x}{3}} } \right)^{10 - r}}{\left( {\frac{{\sqrt 3 }}{{2{x^2}}}} \right)^r} \\ {\text{ = }}{}^{10}{C_r}{\left( {\frac{1}{3}} \right)^{\frac{{10 - r}}{2}}}{\left( {\frac{{\sqrt 3 }}{2}} \right)^r}{\left( x \right)^{\frac{{10 - r}}{2}}}{\left( x \right)^{ - 2r}} \\ {\text{ = }}{}^{10}{C_r}{\left( {\frac{1}{3}} \right)^{\frac{{10 - r}}{2}}}{\left( {\frac{{\sqrt 3 }}{2}} \right)^r}{\left( x \right)^{\frac{{10 - r}}{2} - 2r}} \\\end{aligned} \end{equation} $$

To find the coefficient of term independent of $x$, $$\begin{equation} \begin{aligned} \frac{{10 - r}}{2} - 2r = 0 \\ \therefore r = 2 \\\end{aligned} \end{equation} $$

Therefore, coefficient of term independent of $x$ is

$${}^{10}{C_2}{\left( {\frac{1}{3}} \right)^4}{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = \frac{{9 \times 10}}{2} \times \frac{1}{{3 \times 3 \times 3 \times 3}} \times \frac{3}{4} = \frac{5}{{12}}$$

Question 4. Find the coefficient of ${x^5}$ in ${(1 + 2x)^6}{(1 - x)^7}$.

Solution: We can separately expand both the parts as $$\begin{equation} \begin{aligned} {(1 + 2x)^6} = {}^6{C_0} + {}^6{C_1}(2x) + {}^6{C_2}{(2x)^2} + {}^6{C_3}{(2x)^3} + {}^6{C_4}{(2x)^4} + \\ {}^6{C_5}{(2x)^5} + {}^6{C_6}{(2x)^6} \\ {(1 - x)^7} = {}^7{C_0} - {}^7{C_1}(x) + {}^7{C_2}{( - x)^2} - {}^7{C_3}{(x)^3} + {}^7{C_4}{( - x)^4} - \\ {}^7{C_5}{(x)^5} + {}^7{C_6}{( - x)^6} - {}^7{C_7}{(x)^7} \\\end{aligned} \end{equation} $$

To find the coefficient of ${x^5}$, take the coefficients from both the expansion in such a way that the product of $x$ becomes $5$ i.e.,

$$^6{C_0}({ - ^7}{C_5}){ + ^6}{C_1}(2){(^7}{C_4}){ + ^6}{C_2}{(2)^2}({ - ^7}{C_3}){ + ^6}{C_3}{(2)^3}{(^7}{C_2}){ + ^6}{C_4}{(2)^4}({ - ^7}{C_1}){ + ^6}{C_5}{(2)^5}{(^7}{C_0})$$