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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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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} $

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  • $(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}$
answered Apr 10, 2014 by balaji.thirumalai
edited Apr 10, 2014 by balaji.thirumalai
 

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