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Questions  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Q)

Find the middle term in the expansion of $\big(\large\frac{3}{3}$$+9y\big)^{10}$

$\begin{array}{1 1}(A)\;61236\\(B)\;6236x^2y^4\\(C)\;61236x^5y^5\\(D)\;61236x^4y^3\end{array} $

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A)
Toolbox:
  • If $n+1$ is odd,the middle term is $\large\frac{(n+1)+1}{2}$$+n=\large\frac{n+2}{2}$$+h$
Number of terms in the expansion is $10+1=11$
Middle term of the expansion is $\large\frac{11+1}{2}$$=T_6$
$T_{r+1}=C(10,r)(\large\frac{x}{3})^{10-r}$$(9y)^r$
But $T_{n+r}=T_6$ or $r+1=6$
Putting $r=5$ in (1) we have
$T_6=T_{5+1}=C(10,5)(\large\frac{x}{3})^{10-5}$$(9y)^5$
$\Rightarrow C(10,5)\large\frac{x^5}{3^5}$$.9^5y^5$
$\Rightarrow C(10,5) 3^5 X^5y^5$
$\Rightarrow \large\frac{10!}{5!(10-5)!}$$3^5x^5y^5$
$\Rightarrow \large\frac{10!}{5!5!}$$3^5x^5y^5$
$\Rightarrow \large\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}$$3^5x^5y^5$
$\Rightarrow 61236 x^5y^5$
Hence (C) is the correct answer.
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