Question

Find a, b and n in the expression of ${(a+b)}^n$ if the first,second, three terms of the expansion are 729, 7290 and 30375 respectively.

Answer 1

Its known that ${(r+1)}^{th}$ term, $T_{r+1}$ in the binomial expansion of ${(a+b)}^n$ is

$T_{r+1}={n_C}_r{a}^{n-r}{b}^r$

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

$\therefore$ we obtain

$T_1={n_C}_0{a}^{n-0}{b}^0\Rightarrow a^n=729$ ......$\left(1\right)$

$T_2={n_C}_1{a}^{n-1}{b}^1\Rightarrow n{a}^{n-1}b=7290$ ......$\left(2\right)$

$T_3={n_C}_2{a}^{n-2}{b}^2\Rightarrow \frac{n(n-1)}{2}{a}^{n-2}{b}^2=30375$ ......$\left(3\right)$

Dividing $\left(2\right)$ by $\left(1\right)$

$\Rightarrow \frac{n{a}^{n-1}b}{a^n}=\frac{7290}{729}=10$

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

Dividing $\left(3\right)$ by $\left(2\right)$

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

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

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

$\Rightarrow \frac{b}{a}=10-\frac{25}{3}=\frac{30-25}{3}=\frac{5}{3}$ .......$\left(5\right)$

From $\left(4\right)$ and $\left(5\right)$ we get

$n.\frac{b}{a}=n.\frac{5}{3}=10$

$\Rightarrow n=6$ on simplification

Substituting $n=6$ in eqn$\left(1\right)$ we get $a^6=729$

$\therefore a={729}^{\frac{1}{3}}=3$

From eqn$\left(5\right)$ we obtain

$\frac{b}{3}=\frac{5}{3}\Rightarrow b=5$

Thus, $a=3,b=5$ and $n=6$