Question

The first three terms in the expansion of $(1 + ax)^{n}$ $(n \neq 0)$ are $1$, $6x$ and $16x^{2}$. Then the values of $a$ and $n$ are respectively

• A. 2 and 9
• B. 3 and 2
• C. $\dfrac{2}{3}$ and 9
• D. $\dfrac{3}{2}$ and 6

Hint:

Set up equations from the 2nd and 3rd coefficients of $(1+ax)^n$: $na = \text{coeff of }x$ and $\tfrac{n(n-1)}{2}a^2 = \text{coeff of }x^2$. Divide to eliminate one variable.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Match the binomial expansion coefficients with the given values.
Step 2: Detailed Explanation:
$(1+ax)^n = 1 + nax + \dfrac{n(n-1)}{2}a^2x^2 + \cdots$
$na = 6$ \quad and \quad $\dfrac{n(n-1)}{2}a^2 = 16$.
From $na=6$: $a = 6/n$. Substitute: $\dfrac{n(n-1)}{2}\cdot\dfrac{36}{n^2} = 16 \Rightarrow \dfrac{18(n-1)}{n} = 16 \Rightarrow 18n-18=16n \Rightarrow n=9$, $a=2/3$.
Step 3: Final Answer:
$a = \dfrac{2}{3}$ and $n = 9$.