If for positive integers r 1 and n 2,the | vedclass

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(C) In the expansion of $(1 + x)^{2n}$,the coefficient of $x^k$ is given by $^{2n}C_k$,where $0 \le k \le 2n$.Given that the coefficients of $x^{3r}$

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$n = 2r + 1$

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(C) In the expansion of $(1 + x)^{2n}$,the coefficient of $x^k$ is given by $^{2n}C_k$,where $0 \le k \le 2n$.Given that the coefficients of $x^{3r}$ and $x^{r+2}$ are equal,we have $^{2n}C_{3r} = ^{2n}C_{r+2}$.Using the property $^{n}C_a = ^{n}C_b$,which implies $a = b$ or $a + b = n$,we get two cases:Case $1$: $3r = r + 2$ $\Rightarrow 2r = 2$ $\Rightarrow r = 1$. However,it is given that $r > 1$,so this case is rejected.Case $2$: $3r + (r + 2) = 2n$ $\Rightarrow 4r + 2 = 2n$ $\Rightarrow 2n = 4r + 2$ $\Rightarrow n = 2r + 1$.Thus,the correct relation is $n = 2r + 1$.

If for positive integers $r > 1$ and $n > 2$,the coefficients of the $(3r)^{th}$ and $(r + 2)^{th}$ powers of $x$ in the expansion of $(1 + x)^{2n}$ are equal,then:

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