જો $(a+b)^{n}$ ના વિસ્તરણનાં પ્રથમ ત્રણ પદો અનુક્રમે $729, 7290$ અને $30375$ હોય, તો $a, b$ અને $n$ શોધો.

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 $n=6$