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JEE Main & Advanced JEE Main Paper (Held On 25 April 2013)

  • question_answer
                    If for positive integers\[\operatorname{r}>1,\operatorname{n},>2,\]the coefficients of the \[{{(3r)}^{\operatorname{th}}}\]and \[{{(\operatorname{r}+2)}^{\operatorname{th}}}\]  powers of \[x\]in the expansion of \[{{(1+x)}^{2n}}\] are equal, then n is equal to     JEE Main Online Paper ( Held On 25  April 2013 )

    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) \[LHS:(3r)!(4r+2-3r)!=(3r)!(r+2)!\] \[RHS:(r+2)!(3r)!\]\[\Rightarrow \]\[LHS=RHS\]


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