JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 22)
Let the coefficients of three consecutive terms in the binomial expansion of $$(1+2x)^n$$ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.
Answer
1120
Explanation
$\mathrm{t}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2 \mathrm{x})^{\mathrm{r}}$
$$ \begin{aligned} & \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\ & \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\ & \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{r}+1}}=\frac{5}{8} \\\\ & \Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \\\\ & \Rightarrow 4 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2) \end{aligned} $$
From (1) and (2)
$$ \Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8 $$
$(1) \Rightarrow r=4$
so, coefficient of middle term is
$$ { }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120 $$
$$ \begin{aligned} & \Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2)^{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(2)^{\mathrm{r}}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5} \\\\ & \Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{4}{5} \Rightarrow 5 \mathrm{r}=4 \mathrm{n}-4 \mathrm{r}+4 \\\\ & \Rightarrow 9 \mathrm{r}=4(\mathrm{n}+1) \quad\quad...(1)\\\\ & \Rightarrow \frac{{ }^{n} C_{r}(2)^{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}(2)^{\mathrm{r}+1}}=\frac{5}{8} \\\\ & \Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4} \\\\ & \Rightarrow 4 \mathrm{r}+4=5 \mathrm{n}-5 \mathrm{r} \Rightarrow 5 \mathrm{n}-4=9 \mathrm{r} \quad\quad...(2) \end{aligned} $$
From (1) and (2)
$$ \Rightarrow 4 \mathrm{n}+4=5 \mathrm{n}-4 \Rightarrow \mathrm{n}=8 $$
$(1) \Rightarrow r=4$
so, coefficient of middle term is
$$ { }^{8} \mathrm{C}_{4} 2^{4}=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120 $$
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