Question

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 $n =$

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

5

Answer 2

$\because$ $\frac{T_{2}}{T_{3}} = \frac{2}{n - 2 + 1}.\frac{b}{a} = \frac{2}{n - 1}\left( \frac{b}{a} \right)$ and

$\frac{T_{3}}{T_{4}} = \frac{3}{n + 3 - 3 + 1}.\left( \frac{b}{a} \right) = \frac{3}{n + 1}\left( \frac{b}{a} \right)$

$\because$ $\frac{T_{2}}{T_{3}} = \frac{T_{3}}{T_{4}}$ (given) ;

$\therefore$ $\frac{2}{n - 1}\left( \frac{b}{a} \right) = \frac{3}{n + 1}\left( \frac{b}{a} \right)$ ⇒ $2n + 2 = 3n - 3$ ⇒ $n = 5$