Let the coefficients of three consecutive ter | vedclass

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Question

$\Rightarrow \frac{{ }^n C_{r-1}(2)^{r-1}}{{ }^{ n } C_r(2)^r}=\frac{2}{5}$

$\Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r)

Vedclass · Accepted Answer

$1120$

Vedclass · Answer

$\Rightarrow \frac{{ }^n C_{r-1}(2)^{r-1}}{{ }^{ n } C_r(2)^r}=\frac{2}{5}$

$\Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5}$

$\Rightarrow \frac{r}{n-r+1}=\frac{4}{5} \Rightarrow 5 r=4 n-4 r+4$

$\Rightarrow \frac{{ }^n C_r(2)^r}{{ }^n C_{r+1}(2)^{r+1}}=\frac{5}{8}$

$\Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4}$

$\Rightarrow 4 r+4=5 n-5 r \Rightarrow 5 n-4=9 r \ldots$

From (1) and (2)

$\Rightarrow 4 n +4=5 n -4 \Rightarrow n =8$

$(1)$ $\Rightarrow r=4$

so, coefficient of middle term is

${ }^8 C_4 2^4=16 	imes \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1}=16 	imes 70=1120$

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2 x)^{ n }$ be in the ratio $2: 5: 8$. Then the coefficient of the term, which is in the middle of these three terms, is $...........$.

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