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

Option: 1

3


Option: 2

4


Option: 3

5


Option: 4

6


Answers (1)

best_answer

\ \frac{T_2}{T_3}=\frac{2}{n-2+1} \cdot \frac{b}{a}=\frac{2}{n-1}\left(\frac{b}{a}\right) and \,\frac{T_3}{T_4}=\frac{3}{n+3-3+1} \cdot\left(\frac{b}{a}\right)=\frac{3}{n+1}\left(\frac{b}{a}\right)

 

\frac{T_2}{T_3}=\frac{T_3}{T_4} \quad \text { (given); }\therefore \frac{2}{n-1}\left(\frac{b}{a}\right)=\frac{3}{n+1}\left(\frac{b}{a}\right) \Rightarrow 2 n+2=3 n-3 \Rightarrow n=5

 

 

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