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The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1 : 4 are:$[\text{Hint}:\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}\ \frac{\text{r}+1}{24-\text{r}}\ \frac{1}{4}\Rightarrow4\text{r}+4=24-4\Rightarrow\text{r}=4]$

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The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1 : 4 are:$[\text{Hint}:\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}\ \frac{\text{r}+1}{24-\text{r}}\ \frac{1}{4}\Rightarrow4\text{r}+4=24-4\Rightarrow\text{r}=4]$

Aicharya · Accepted Answer

Solution: Let the two successive terms in the expansion of (1 + x) 24 be (r + 1)(r + 2) th terms. Now, $\text{T}_{\text{r}+1}=\ ^{24}\text{C}_\text{r}\text{x}^\text{r}\ \text{and}\ \text{T}_{\text{r}+2}=\ ^{24}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$ Given that, $\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}$ $\Rightarrow\frac{\frac{(24)!}{\text{r}!(24-\text{r})!}}{\frac{(24)!}{(\text{r}+1)!(24-\text{r}-1)!}}=\frac{1}{4}$ $\Rightarrow\frac{(\text{r}+1)\text{r}!(23-\text{r})!}{\text{r}!(24-\text{r})(23-\text{r})!}=\frac{1}{4}\Rightarrow\frac{\text{r}+1}{24-\text{r}}=\frac{1}{4}$ $\Rightarrow4\text{r}+4=24-\text{r}\Rightarrow\text{r}=4$ $\therefore\text{T}_{4+1}=\text{T}_5\ \text{and}\ \text{T}_{4+2}=\text{T}_6$ Hence. 5 th and 6 th terms.

Choose the correct answer. The two successive terms in the expansion of (1 + x) 24 whose coefficients are in…

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