If frac{T_2}{T_3} in the expansion of (a + b) | vedclass

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(C) The general term in the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$. For the expansion of $(a + b)^n$,we have $\frac{T_2}{T_3} = \fr

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$5$

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(C) The general term in the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$. For the expansion of $(a + b)^n$,we have $\frac{T_2}{T_3} = \frac{^nC_1 a^{n-1} b}{^nC_2 a^{n-2} b^2} = \frac{n}{\frac{n(n-1)}{2} \cdot \frac{b}{a}} = \frac{2}{n-1} \cdot \frac{a}{b}$. For the expansion of $(a + b)^{n+3}$,we have $\frac{T_3}{T_4} = \frac{^{n+3}C_2 a^{n+1} b^2}{^{n+3}C_3 a^n b^3} = \frac{\frac{(n+3)(n+2)}{2}}{\frac{(n+3)(n+2)(n+1)}{6}} \cdot \frac{a}{b} = \frac{3}{n+1} \cdot \frac{a}{b}$. Given that $\frac{T_2}{T_3} = \frac{T_3}{T_4}$,we equate the expressions: $\frac{2}{n-1} \cdot \frac{a}{b} = \frac{3}{n+1} \cdot \frac{a}{b}$. This simplifies to $\frac{2}{n-1} = \frac{3}{n+1}$. Cross-multiplying gives $2(n + 1) = 3(n - 1)$,which implies $2n + 2 = 3n - 3$. Solving for $n$,we get $n = 5$.

If $\frac{T_2}{T_3}$ in the expansion of $(a + b)^n$ and $\frac{T_3}{T_4}$ in the expansion of $(a + b)^{n + 3}$ are equal,then $n=$

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