Question

If $(1 + ax)^{n} = 1 + 8x + 24x^{2} + .....$ then the value of a and n is

• A. $2,4$
• B. 2, 3
• C. 3, 6
• D. 1, 2

Answer 1

Answer: (A)

$2,4$

Answer 2

We know that $(1 + x)^{n} = 1 + \frac{nx}{1!} + \frac{n(n - 1)x^{2}}{2!} + .....$

$(1 + ax)^{n} = 1 + \frac{n(ax)}{1!} + \frac{n(n - 1)(ax)^{2}}{2!} + ......$

⇒$1 + 8x + 24x^{2} + ...... = 1 + \frac{n(ax)}{1!} + \frac{n(n - 1)(ax)^{2}}{2!} + ....$Comparing coefficients of both sides we get, $na = 8,$ and $\frac{n(n - 1)a^{2}}{2!} = 24$ on solving, $a = 2$, b = 4.`