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In the binomial expansion of $(1+ax)^n$, the first three terms are $1+12x+64x^2$, then $n=?$

$\begin{array}{1 1} 6 \\ 9 \\ 10 \\ 12 \end{array} $

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Given first three terms in the expansion $(1+ax)^n$ are $ 1,12x,64x^2$
$\Rightarrow\:^nC_1.a=12\:\:and\:\:^nC_2.a^2=64$
$\Rightarrow\:na=12\:\:and\:\:\large\frac{n(n-1)}{2}$$.a^2=64$
$\Rightarrow\:na(na-a)=128$
$\Rightarrow\:12(12-a)=128$ or $a=\large\frac{4}{3}$
$\Rightarrow\:n=9$
answered Sep 26, 2013 by rvidyagovindarajan_1
 

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