The first 3 terms in the expansion of (1 + ax | vedclass

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(C) The binomial expansion of $(1 + ax)^n$ is given by $1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots$Given the first three terms are $1, 6x, 16x^2$,we h

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$2/3$ and $9$

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(C) The binomial expansion of $(1 + ax)^n$ is given by $1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots$Given the first three terms are $1, 6x, 16x^2$,we have:$n(ax) = 6x \implies na = 6$ $(i)$$\frac{n(n-1)}{2} a^2 x^2 = 16x^2 \implies n(n-1)a^2 = 32$ $(ii)$From $(i)$,$a = \frac{6}{n}$. Substituting this into $(ii)$:$n(n-1) \left(\frac{6}{n}ight)^2 = 32$$n(n-1) \frac{36}{n^2} = 32$$\frac{n-1}{n} 	imes 36 = 32$$\frac{n-1}{n} = \frac{32}{36} = \frac{8}{9}$$9n - 9 = 8n \implies n = 9$Substituting $n = 9$ into $(i)$,$9a = 6 \implies a = \frac{6}{9} = \frac{2}{3}$.Thus,$a = 2/3$ and $n = 9$.

The first $3$ terms in the expansion of $(1 + ax)^n$ $(n \ne 0)$ are $1, 6x$ and $16x^2$. Then the value of $a$ and $n$ are respectively

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