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If the integers $r>1, n>2$ and coefficients of $(3 r)^{\mathrm{th}}$ and $(\mathrm{r}+2)^{\text {nd }}$ terms in binomial expansion of $(1+\mathrm{x})^{2 \mathrm{n}}$ are equal, then
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The correct answer is:
$\mathrm{n}=2 \mathrm{r}$
$\mathrm{n}=2 \mathrm{r}$
$\mathrm{t}_{r+2}={ }^{2 n} C_{r+1} x^{r+1} ; t_{3 r}={ }^{2 n} C_{3 r-1} x^{3 r-1}$
Given ${ }^{2 n} \mathrm{C}_{r+1}={ }^{2 n} \mathrm{C}_{3 r-1} ; \Rightarrow{ }^{2 n} \mathrm{C}_{2 n-(r+1)}={ }^{2 n} \mathrm{C}_{3 r-1}$ $\Rightarrow 2 n-r-1=3 r-1 \Rightarrow 2 n=4 r \Rightarrow n=2 r$
Given ${ }^{2 n} \mathrm{C}_{r+1}={ }^{2 n} \mathrm{C}_{3 r-1} ; \Rightarrow{ }^{2 n} \mathrm{C}_{2 n-(r+1)}={ }^{2 n} \mathrm{C}_{3 r-1}$ $\Rightarrow 2 n-r-1=3 r-1 \Rightarrow 2 n=4 r \Rightarrow n=2 r$
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