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If \frac{T_2}{T_3} in the expansion of (a + b)n and \frac{T_3}{T_4} in the expansion of \left(\begin{array}{ll} a + b \end{array}\right)^{n+3} are equal, then n is equal to

 

Option: 1

3


Option: 2

4


Option: 3

5


Option: 4

6


Answers (1)

best_answer

\frac{{ }^n C_1 a^{n-1} b}{{ }^n C_2 a^{n-2} b^2}=\frac{{ }^{n+3} C_2 a^{n+1} b^2}{{ }^{n+3} C_3 a^n b^3} \Rightarrow \frac{n}{n(n-1)}=\frac{(n+3)(n+2)}{2} \frac{2}{(n+3)(n+2)(n+1)}

\Rightarrow \frac{2}{n-1}=\frac{3}{n+1} \Rightarrow 2 n+2=3 n-3 \Rightarrow n=5

Hence (C) is the correct answer.

Posted by

Pankaj

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