Solve the following


If $T_{2} / T_{3}$ in the expansion of $(a+b)^{n}$ and $T_{3} / T_{4}$ in the expansion of $(a+b)^{n+3}$ are equal, then $n=$

(a) 3

(b) 4

(c) 5

(d) 6


(c) 5

In the expansion $(a+b)^{n}$, we have

$\frac{T_{2}}{T_{3}}=\frac{{ }^{n} C_{1} a^{n-1} \times b^{1}}{{ }^{n} C_{2} a^{n-2} \times b^{2}}$

In the expansion $(a+b)^{n+3}$, we have

$\frac{T_{3}}{T_{4}}=\frac{{ }^{n+3} C_{2} a^{n+1} b^{2}}{{ }^{n+3} C_{3} a^{n} b^{3}}$

Thus, we have


$\Rightarrow \frac{{ }^{n} C_{1} a}{{ }^{n} C_{2} b}=\frac{{ }^{n+3} C_{2} a}{{ }^{n+3} C_{3} b}$

$\Rightarrow \frac{2}{n-1}=\frac{3}{n+1}$

$\Rightarrow 2 n+2=3 n-3$

$\Rightarrow n=5$


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