If $\frac{\text{T}_{2}}{\text{T}_{3}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}}$ and $\frac{\text{T}_{3}}{\text{T}_{4}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}+3}$ are equal, then n =

Question

Aicharya · Accepted Answer

Solution: In the expansion $(\text{a}+\text{b})^{\text{n}},$ we have $\frac{\text{T}_{2}}{\text{T}_{3}}=\frac{{^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}\times\text{b}^{1}}{{^\text{n}}\text{C}_{\text{2}}\text{a}^{\text{n}-2}\times\text{b}^{2}}$ In the expansion $(\text{a}+\text{b})^{\text{n}+3},$ we have $\frac{\text{T}_{3}}{\text{T}_{4}}=\frac{{^\text{n+3}}\text{C}_{\text{2}}\text{a}^{\text{n}+1}\times\text{b}^{2}}{{^\text{n+3}}\text{C}_{\text{3}}\text{a}^{\text{n}}\times\text{b}^{3}}$ Thus, we have $\frac{\text{T}_{2}}{\text{T}_{3}}=\frac{\text{T}_{3}}{\text{T}_{4}}$ $\Rightarrow \frac{{^\text{n}}\text{C}_{\text{1}}\ \text{a}}{{^\text{n}}\text{C}_{\text{2}}\ \text{b}}=\frac{{^\text{n+3}}\text{C}_{\text{2}}\ \text{a}}{{^\text{n+3}}\text{C}_{\text{3}}\ \text{b}}$ $\Rightarrow \frac{2}{\text{n}-1}=\frac{3}{\text{n}+1}$ $\Rightarrow 2\text{n}+2=3\text{n}-3$ $\Rightarrow \text{n}=5$

If \(\frac{\text{T}_{2}}{\text{T}_{3}}\) in the expansion of \((\text{a}+\text{b})^{\text{n}}\) and…

Solution