If for positive integers r 1, n 2, t | vedclass

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Question

Expansion of $(1+x)^{2 n}$ is $1+^{2 n} C_{1} x+^{2 n} C_{2} x^{2}$

$+\ldots \ldots+^{2 n} C_{r} x^{r}+^{2 n} C_{r+1} x^{r+1}+\ldots \ldots+^{2 n}

Vedclass · Accepted Answer

$2r+ 1$

Vedclass · Answer

Expansion of $(1+x)^{2 n}$ is $1+^{2 n} C_{1} x+^{2 n} C_{2} x^{2}$

$+\ldots \ldots+^{2 n} C_{r} x^{r}+^{2 n} C_{r+1} x^{r+1}+\ldots \ldots+^{2 n} C_{2 n} x^{2 n}$

As given $^{2n}{{	ext{C}}_{r + 2}}{ = ^{2n}}{{	ext{C}}_{3r}}$

$ \Rightarrow \frac{{(2n)!}}{{(r + 2)!(2n - r - 2)!}} = \frac{{(2n)!}}{{(3r)!(2n - 3r)!}}$

$\Rightarrow(3 r) !(2 n-3 r) !=(r+2) !(2 n-r-2) !.........(1)$

Now, put value of $n$ from the given choices.

Choice $(a)$ put $n=2 r+1$ in $(1)$

$LHS:$ $(3 r) !(4 r+2-3 r) !=(3 r) !(r+2) !$

$\mathrm{RHS}:(r+2) !(3 r) !$

$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

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

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