If (1 + ax)^n = 1 + 8x + 24x^2 + ...., then t | vedclass

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Question

(A) Given the binomial expansion: $(1 + ax)^n = 1 + n(ax) + \frac{n(n - 1)}{2}(ax)^2 + .... = 1 + 8x + 24x^2 + ....$Comparing the coefficients of $x$

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$2, 4$

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(A) Given the binomial expansion: $(1 + ax)^n = 1 + n(ax) + \frac{n(n - 1)}{2}(ax)^2 + .... = 1 + 8x + 24x^2 + ....$Comparing the coefficients of $x$ and $x^2$:$na = 8$  --- $(1)$$\frac{n(n - 1)}{2}a^2 = 24$  --- $(2)$From $(1)$,$n = \frac{8}{a}$. Substituting this into $(2)$:$\frac{(\frac{8}{a})(\frac{8}{a} - 1)}{2}a^2 = 24$$\frac{8}{2a} 	imes (8 - a) 	imes a = 24$$4(8 - a) = 24$$8 - a = 6$$a = 2$Substituting $a = 2$ into $(1)$:$n(2) = 8 \Rightarrow n = 4$.Thus,the values are $a = 2$ and $n = 4$.

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

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