A) \[2r+1\]
B) \[2r-1\]
C) 3r
D) \[r+1\]
Correct Answer: A
Solution :
Expansion of \[{{(1+x)}^{2n}}\]is\[1{{+}^{2n}}{{C}_{1}}x{{+}^{2n}}{{C}_{2}}{{x}^{2}}\] \[+......{{+}^{2n}}{{C}_{r}}{{x}^{r}}{{+}^{2n}}{{C}_{r+1}}{{x}^{r+1}}+......{{+}^{2n}}{{C}_{2n}}{{x}^{2n}}\] As given \[^{2n}{{C}_{r+2}}{{=}^{2n}}{{C}_{3r}}\] \[\Rightarrow \]\[\frac{(2n)!}{(r+2)!(2n-r-2)!}=\frac{(2n)!}{(3r)!(2n-3r)!}\] \[\Rightarrow \]\[(r+2)!(2n-3r)!=(r+2)!(2n-r-2)!\]?(1) Now, put value of n from the given choices. Choice (a) put \[n=2r+1\] in (1) \[LHS:(3r)!(4r+2-3r)!=(3r)!(r+2)!\] \[RHS:(r+2)!(3r)!\]\[\Rightarrow \]\[LHS=RHS\]You need to login to perform this action.
You will be redirected in
3 sec