# Find the middle terms in the expansions of $\;\large ($$3$$- \large\frac{x^3}{6})^7$

$\begin{array}{1 1} T_4 = -\large\frac{105}{8}x^9 \qquad T_5 = \large\frac{35}{48}x^{12} \\ T_4 = \large\frac{105}{8}x^9 \qquad T_5 = \large\frac{35}{48}x^{12} \\ T_4 = \large\frac{105}{8}x^9 \qquad T_5 = -\large\frac{35}{48}x^{12} \\ T_4 = -\large\frac{105}{8}x^9 \qquad T_5 = -\large\frac{35}{48}x^{12}\end{array}$

Toolbox:
• $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n, where b^0 = 1 = a^{n-n} • Looking at the pattern of the successive terms we can say that the (r + 1)\text{th} term T_{r+1} = ^n\large C$$_r\;a^{n–r}b^r$. This is also called the general term of the expansion.
• If $n$ is odd, then $n +1$ is even, so there will be two middle terms in the expansion of $(a+b)^n$ which are $\large\frac{n+1}{2}$$^{\text{th}} term and \large\frac{n+1}{2}$$+1$$^{\text{th}} term Given n = 7, the middle terms are \large\frac{7+1}{2}$$^{\text{th}}$ and $\large\frac{7+1}{2} $$+1$$^{\text{th}}$ terms, i.e, $4^{\text{th}}$ and $5^{\text{th}}$ terms.
Given $\;\large ($$3$$- \large\frac{x^3}{6})^7$, $n=7, r = 3$ or $4, x = 3, y = \large\frac{-x^3}{6}$
$\Rightarrow T_4 = T_{3+1} = \; ^7\large C$$_3 3^{7-3} (\large\frac{-x^3}{6})^{\normalsize 3} \Rightarrow T_4 = T_{3+1} = -1^3 \;^7\large C$$_3 3^4 \large\frac{x^9}{6^3}$
$\Rightarrow T_4 = -\large\frac{105}{8}$$x^9 \Rightarrow T_5 = T_{4+1} = \; ^7\large C$$_4 3^{7-4} (\large\frac{-x^3}{6})^{\normalsize 4}$
$\Rightarrow T_5 = -1^4 \;^7\large C$$_4 \;3^3 \large\frac{x^{12}}{6^4} \Rightarrow T_5 = \large\frac{35}{48}$$x^{12}$
edited Apr 10, 2014