Find $a, b$ and $n$ in the expansion of $(a+b)^{n}$ if the first three terms of the expansion are $729,7290$ and $30375,$ respectively.
It is known that $(r+1)^{th}$ term, $\left(T_{r+1}\right),$ in the binomial expansion of $(a+b)^{n}$ is given by
${T_{r + 1}} = {\,^n}{C_r}{a^{n - r}}{b^r}$
The first three terms of the expansion are given as $729,7290$ and $30375$ respectively.
Therefore, we obtain
${T_1} = {\,^n}{C_0}{a^{n - 0}}{b^0} = {a^n} = 729$ ........$(1)$
${T_2} = {\,^n}{C_1}{a^{n - 1}}{b^1} = n{a^{n - 1}}b = 7290$ ...........$(2)$
${T_2} = {\,^n}{C_1}{a^{n - 2}}{b^2} = \frac{{n(n - 1)}}{2}{a^{n - 2}}{b^2} = 30375$ ..........$(3)$
Diving $(2)$ and $(1),$ we obtain
$\frac{n a^{n-1} b}{a^{n}}=\frac{7290}{729}$
$\Rightarrow \frac{n b}{a}=10$ ..........$(4)$
Dividing $(3)$ by $(2),$ we obtain
$\frac{n(n-1) a^{n-2} b^{2}}{2 n a^{n-1} b}=\frac{30375}{7290}$
$\Rightarrow \frac{(n-1) b}{2 a}=\frac{30375}{7290}$
$\Rightarrow \frac{(n-1) b}{a}=\frac{30375 \times 2}{7290}=\frac{25}{3}$
$\Rightarrow \frac{n b}{a}-\frac{b}{a}=\frac{25}{3}$
$ \Rightarrow 10 - \frac{b}{a} = \frac{{25}}{3}\quad $ [ Using $(1)$ ]
$\Rightarrow \frac{b}{a}=10-\frac{25}{3}=\frac{5}{3}$ ............$(5)$
From $(4)$ and $(5),$ we obtain
$n \cdot \frac{5}{3}=10$
$\Rightarrow n=6$
Substituting $n=6$ in equation $(1),$ we obtain $a^{6}$
$=729$
$\Rightarrow a=\sqrt[6]{729}=3$
From $(5),$ we obtain
$\frac{b}{3}=\frac{5}{3} \Rightarrow b=5$
Thus, $a=3, b=5,$ and $n=6$
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