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

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Solution

(c) 5
$In the expansion (a + b)^{n}, we have \frac{T_{2}}{T_{3}} = \frac{{}^{n}C_{1} a^{n - 1} \times b^{1}}{{}^{n}C_{2} a^{n - 2} \times b^{2}} In the expansion (a + b)^{n + 3}, we have \frac{T_{3}}{T_{4}} = \frac{{}^{n + 3}C_{2} a^{n + 1} b^{2}}{{}^{n + 3}C_{3} a^{n} b^{3}} Thus, we have \frac{T_{2}}{T_{3}} = \frac{T_{3}}{T_{4}} \Rightarrow \frac{{}^{n}C_{1} a}{{}^{n}C_{2} b} = \frac{{}^{n + 3}C_{2} a}{{}^{n + 3}C_{3} b} \Rightarrow \frac{2}{n - 1} = \frac{3}{n + 1} \Rightarrow 2 n + 2 = 3 n - 3 \Rightarrow n = 5$

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