Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729,

Solution:

It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $\mathrm{T}_{r+1}={ }^{n} \mathrm{C}_{r} \mathrm{a}^{n-t} \mathrm{~b}^{t}$.

The first three terms of the expansion are given as 729,7290 , and 30375 respectively.

Therefore, we obtain

$T_{1}={ }^{n} C_{0} a^{n-0} b^{0}=a^{n}=729$  $\ldots(1)$

$T_{2}={ }^{n} C_{1} a^{n-1} b^{\prime}=n a^{n-1} b=7290$  $\ldots(2)$

$T_{3}={ }^{n} C_{2} a^{n-2} b^{2}=\frac{n(n-1)}{2} a^{n-2} b^{2}=30375$   ..(3)

Dividing $(2)$ by $(1)$, we obtain

$\frac{\mathrm{na}^{\mathrm{n}-1} \mathrm{~b}}{\mathrm{a}^{\mathrm{n}}}=\frac{7290}{729}$

$\Rightarrow \frac{\mathrm{nb}}{\mathrm{a}}=10$  ...(4)

Dividing $(3)$ by $(2)$, we obtain

$\frac{\mathrm{n}(\mathrm{n}-1) \mathrm{a}^{\mathrm{n}-2} \mathrm{~b}^{2}}{2 \mathrm{n} \mathrm{a}^{\mathrm{n}-1} \mathrm{~b}}=\frac{30375}{7290}$

$\Rightarrow \frac{(\mathrm{n}-1) \mathrm{b}}{2 \mathrm{a}}=\frac{30375}{7290}$

$\Rightarrow \frac{(\mathrm{n}-1) \mathrm{b}}{\mathrm{a}}=\frac{30375 \times 2}{7290}=\frac{25}{3}$

$\Rightarrow \frac{\mathrm{nb}}{\mathrm{a}}-\frac{\mathrm{b}}{\mathrm{a}}=\frac{25}{3}$

$\Rightarrow 10-\frac{\mathrm{b}}{\mathrm{a}}=\frac{25}{3}$ $[U \operatorname{sing}(4)]$

$\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=10-\frac{25}{3}=\frac{5}{3}$ $\ldots(5)$

From (4) and (5), we obtain

$\mathrm{n} \cdot \frac{5}{3}=10$

$\Rightarrow \mathrm{n}=6$

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 $n=6$.