A) \[n=2r\]
B) \[n=2r+1\]
C) \[n=3r\]
D) None of these
Correct Answer: A
Solution :
\[3r\,\,\text{th}\] term in the expansion of\[{{(1+x)}^{2n}}\] \[{{=}^{2n}}{{C}_{3r-1}}{{x}^{3r-1}}\] and\[(r+2)\text{th}\]term in the expansion of\[{{(1+x)}^{2n}}\] \[{{=}^{2n}}{{C}_{r+1}}{{x}^{r+1}}\] Given that the binomial coefficients of\[(3r)th\]and\[(r+2)th\]terms are equal. \[\therefore \] \[^{2n}{{C}_{3r-1}}{{=}^{2n}}{{C}_{r+1}}\] \[\Rightarrow \] \[3r-1=r+1\] or \[2n=(3r-1)+(r+1)\] \[\Rightarrow \] \[2r=2\] or \[2n=4r\] \[\Rightarrow \] \[r=1\] or \[n=2r\] But \[r>1\] \[n=2r\]You need to login to perform this action.
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