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Question: If \(\frac{T_{2}}{T_{3}}\)is expansion of (a + b)<sup>n</sup> = \(\frac{T_{3}}{T_{4}}\)in expansion ...

If T2T3\frac{T_{2}}{T_{3}}is expansion of (a + b)n = T3T4\frac{T_{3}}{T_{4}}in expansion of (a + b)n + 3 then n=

A

3

B

4

C

5

D

None

Answer

5

Explanation

Solution

nC1.an−1b1nC2.an−2b2\frac{nC_{1}.a^{n - 1}b^{1}}{nC_{2}.a^{n - 2}b^{2}}=n+3C2.an+1b2n+3C3.anb3\frac{n + 3C_{2}.a^{n + 1}b^{2}}{n + 3C_{3}.a^{n}b^{3}} (check option)

ŽnC2nC1\frac{nC_{2}}{nC_{1}}=n+3C3n+3C2\frac{n + 3C_{3}}{n + 3C_{2}}

Žn−12\frac{n - 1}{2}= n+13\frac{n + 1}{3}

Ž 3n –3 = 2n + 1

n = 5