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Question

If the coefficient of $$(2r+4)^{th}$$ term and $$(r-2)^{th}$$ term in the expansion of $$(1+x)^{18}$$ are equal then $$r$$ 


A
9
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B
4
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C
6
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D
3
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Solution

The correct option is C $$6$$
In the expansion of $$(1+x)^{18}$$
$$T_{r+1}=^{18}C_{r}1^{18-r}x^r$$
$$\therefore T_{2r+4}=^{18}C_{2r+3}1^{18-2r-3}x^{2r+3}$$
and $$T_{r-2}=^{18}C_{r-3}1^{18-r+3}x^{r-3}$$
$$ T_{2r+4}=T_{r-2}$$
i.e.$$^{18}C_{2r+3}=^{18}C_{r-3}$$
$$\dfrac{18!}{(18-2r-3)!(2r+3)!}=\dfrac{18!}{(18-r+3)!(r-3)!}$$
$$(18-2r-3)!(2r+3)!=(18-r+3)!(r-3)!$$
$$(15-2r)!(2r+3)!=(21-r)!(r-3)!$$
Now, Posibilities are $$15-2r=21-r\Rightarrow r=-6$$
But we know $$r$$ is a an integer such that $$r\space\epsilon\space[0,18]$$
Now, $$15-2r=r-3\Rightarrow 3r=18\Rightarrow r=6$$
Checking with $$2r+3=21-r\Rightarrow 3r=18 \Rightarrow r=6$$

Mathematics

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