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QJEEM23Ma2841: JEE Main 2023 Mathematics - Class 11 JEEM23Ma2841 Solution: Let the coefficients of three consecutive terms in the binomial …

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

Step-by-Step Verified Solution

tr+1=nCr(2x)r\mathrm{t}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2 \mathrm{x})^{\mathrm{r}}

amp;⇒nCr−1(2)r−1nCr(2)r=25amp;⇒n!(r−1)!(n−r+1)!n!(2)r!(n−r)!=25amp;⇒rn−r+1=45⇒5r=4n−4r+4amp;⇒9r=4(n+1)...(1)amp;⇒nCr(2)rnCr+1(2)r+1=58amp;⇒n!r!(n−r)!n!(r+1)!(n−r−1)!=54⇒r+1n−r=54amp;⇒4r+4=5n−5r⇒5n−4=9r...(2)\begin{aligned}& \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\& \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\& \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\& \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\& \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{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 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2)\end{aligned}

From (1) and (2)

⇒4n+4=5n−4⇒n=8\Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8

(1)⇒r=4(1) \Rightarrow r=4

so, coefficient of middle term is

8C424=16×8×7×6×54×3×2×1=16×70=1120{ }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120

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