Binomial Theorem 1 Question 27
29. If $(1+a x)^{n}=1+8 x+24 x^{2}+\ldots$, then $a=\ldots$ and $n=\ldots$.
$(1983,2 M)$
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Solution:
- Given,
$$ \begin{aligned} & \quad(1+a x)^{n}=1+8 x+24 x^{2}+\ldots \\ & \Rightarrow \quad 1+a n x+\frac{n(n-1)}{2 !} a^{2} x^{2}+\ldots=1+8 x+24 x^{2}+\ldots \\ & \therefore \quad a n=8 \text { and } a^{2} \frac{n(n-1)}{2}=24 \\ & \Rightarrow \quad 8(8-a)=48 \\ & \Rightarrow \quad 8-a=6 \quad \Rightarrow \quad a=2 \\ & \text { Hence, } \quad a=2 \quad \text { and } \quad n=4 \end{aligned} $$
