Binomial Theorem
6.0 Middle Term
6.0 Middle Term
(i) If $n$ is even, total number of terms in the expansion of ${(x + y)^n}$ is $n+1$ i.e., odd. So, there is only one midldle term i.e., ${\left( {\frac{n}{2} + 1} \right)^{th}}$ term. In general form, it is written as $${T_{\frac{n}{2} + 1}} = {}^n{C_{\frac{n}{2}}}{x^{\frac{n}{2}}}{y^{\frac{n}{2}}}$$
(ii) If $n$ is odd, total number of terms in the expansion of ${(x + y)^n}$ is $n+1$ i.e., even. So, there are two middle terms i.e., ${\left( {\frac{{n + 1}}{2}} \right)^{th}}$ term and ${\left( {\frac{{n + 3}}{2}} \right)^{th}}$ term. In general form, they are written as $$^n{C_{\frac{{n - 1}}{2}}}{x^{\frac{{n + 1}}{2}}}{y^{\frac{{n - 1}}{2}}}{\text{ and}}{{\text{ }}^n}{C_{\frac{{n + 1}}{2}}}{x^{\frac{{n - 1}}{2}}}{y^{\frac{{n + 1}}{2}}}$$
Question 8. Find the middle term in the expansion of $${\left( {3x - \frac{{{x^3}}}{6}} \right)^7}$$
Solution: As $n$ is odd, there are two midle terms i.e., $$\begin{equation} \begin{aligned} \frac{{n + 1}}{2} = \frac{{7 + 1}}{2} = {4^{th}}{\text{ term and}} \\ \frac{{n + 3}}{2} = \frac{{7 + 3}}{2} = {5^{th}}{\text{ term}} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} {T_4} = {}^7{C_3}{(3x)^4}{\left( { - \frac{{{x^3}}}{6}} \right)^3} \\ {\text{ = }} - {}^7{C_3}{.3^4}{x^4}\frac{{{x^9}}}{{{6^3}}} \\ {\text{ = }} - \frac{{7!}}{{4!3!}}{x^{13}}\frac{{3 \times 3 \times 3 \times 3}}{{6 \times 6 \times 6}} \\ {\text{ = }} - \frac{{105}}{8}{x^{13}} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} {T_5} = {}^7{C_4}{(3x)^3}{\left( { - \frac{{{x^3}}}{6}} \right)^4} \\ {\text{ = }}{}^7{C_4}{.3^3}{x^3}\frac{{{x^{12}}}}{{{6^4}}} \\ {\text{ = }}35{x^{15}}\frac{{3 \times 3 \times 3}}{{6 \times 6 \times 6 \times 6}} \\ {\text{ = }}\frac{{35}}{{48}}{x^{15}} \\\end{aligned} \end{equation} $$
