Browse Questions

# Find the middle terms in the expansions of $\; \large ( \frac{x}{3}$$+9y \large)$$^{10}$

$\begin{array}{1 1} 61236 x^5y^5 \\ 61236 x^5y^4 \\ 61236 x^4y^5 \\ 61236 x^6y^6\end{array}$

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 even, there will be only one middle term in the expansion of $(a+b)^n$ which is $\large\frac{n}{2} $$+1$$^{\text{th}}$ term
Given $n=10$ is even, there will be only one middle term in the expansion which is $\large\frac{10}{2} $$+1$$^{\text{th}}$$= 6^{\text{th}} term Given \; \large ( \frac{x}{3}$$+9y \large)$$^{10},$$\; n = 10, r = 5, a = \large\frac{x}{3}$$, b = 9y \Rightarrow T_6 = T_{5+1} = \; ^{10}\large C$$_5 (\large\frac{x}{3})$$^{10-5} (9y)^{\normalsize 5} \Rightarrow T_6 = \;^{10} \large C$$_5\; x^5 y^5 \large\frac{9^5}{3^5}$
$\Rightarrow T_6 = 61236 x^5y^5$
edited Apr 10, 2014