JEE Main 2023 — Mathematics Algebra

Given,

The coefficients of three consecutive terms in the binomial expansion of $(1+2x{)}^{n}$(1+2x)n be in the ratio $2:5:8$2:5:8.

Now ${r}^{th}$rth term in expansion of $(1+2x{)}^{n}$(1+2x)n is given by, ${t}_{r+1}=Crn(2x{)}^{r}$tr+1​=Crn(2x)r

Now let ${T}_{r},{T}_{r+1},{T}_{r+2}$Tr​,Tr+1​,Tr+2​ are in the ratio $2:5:8$2:5:8

$$\Rightarrow \frac{{T}_{r}}{{T}_{r+1}}=\frac{Cr-1n(2{)}^{r-1}}{Crn(2{)}^{r}}=\frac{2}{5}$$⇒Tr+1​Tr​​=Crn(2)rCr−1n(2)r−1​=52​

$$\Rightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!(2)}{r!(n-r)!}}=\frac{2}{5}$$⇒r!(n−r)!n!(2)​(r−1)!(n−r+1)!n!​​=52​

$$\Rightarrow \frac{r}{n-r+1}=\frac{4}{5}\Rightarrow 5r=4n-4r+4$$⇒n−r+1r​=54​⇒5r=4n−4r+4

$$\Rightarrow 9r=4(n+1).......(1)$$⇒9r=4(n+1).......(1)

Now taking other ratio $\frac{{T}_{r+1}}{{T}_{r+2}}=\frac{Crn(2{)}^{r}}{Cr+1n(2{)}^{r+1}}=\frac{5}{8}$Tr+2​Tr+1​​=Cr+1n(2)r+1Crn(2)r​=85​

$$\Rightarrow \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{5}{4}\Rightarrow \frac{r+1}{n-r}=\frac{5}{4}$$⇒(r+1)!(n−r−1)!n!​r!(n−r)!n!​​=45​⇒n−rr+1​=45​

$$\Rightarrow 4r+4=5n-5r$$⇒4r+4=5n−5r

$$\Rightarrow 5n-4=9r.........(2)$$⇒5n−4=9r.........(2)

From equation (1)&(2) we get,

$4n+4=5n-4\Rightarrow n=8$4n+4=5n−4⇒n=8 and $r=4$r=4

So, coefficient of middle term will be

$$C48{2}^{4}=16\times \frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}=16\times 70=1120$$C4824=16×4×3×2×18×7×6×5​=16×70=1120