Question

In the expansion of (1+ ax)ⁿ, the first three terms are respectively 1, 12x and 64x². What is n equal to?
1. 6
2. 9
3. 10
4. 12

Answer 1

Correct Answer - Option 2 : 9

Concept:

The general term in the binomial expansion of (a + b)ⁿ is given by: \({T_{r + 1}} = {\;^n}{C_r} \times {a^{n - r}} \times {b^r}\)

Calculation:

Given:

The first three terms are respectively 1, 12x and 64x².

2^st term in expansion of (1+ ax)ⁿ  = \(^n{C_1} \times {\left( 1 \right)^{n - 1}} \times {\left( {ax} \right)^1}\)

⇒ \(^n{C_1} \times {\left( 1 \right)^{n - 1}} \times {\left( {ax} \right)^1} = n \times a \times x = 12\;x\)

⇒ n × a = 12

⇒ a = 12/n

3^rd term in expansion of (1+ ax)ⁿ = \(^n{C_2} \times {\left( 1 \right)^{n - 2}} \times {\left( {ax} \right)^2}\)

⇒ \(^n{C_2} \times {\left( 1 \right)^{n - 2}} \times {\left( {ax} \right)^2} = \frac{{n \times \left( {n - 1} \right)}}{2} \times {a^2} \times {x^2} = 64\;{x^2}\)

\(\Rightarrow \;\frac{{n \times \left( {n - 1} \right)}}{2} \times {a^2} = 64\)     .....(1)

By substituting the value of a in equation (1), we get

\( \Rightarrow \;\frac{{n \times \left( {n - 1} \right)}}{2} \times \frac{{144}}{{{n^2}}} = 64\)

⇒ 8n = 72

⇒ n = 9

∴ The value of n is 9.