Question

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.

Answer 1

We have,

$T_1=n_{C_0}{a}^n{b}^0=729$ ...(i)

$T_2=n_{C_1}{a}^{n-1}b=7290$ ...(ii)

$T_3=n_{C_2}{a}^{n-2}{b}^2=30375$ ...(iii)

From (i) $a^n=729$ ...(iv)

From (ii) $n{a}^{n-1}b=7290$ ...(v)

From (iii) $\frac{n(n-1)}{2}{a}^{n-2}{b}^2=30375$ ...(vi)

Multiplying (iv) and (vi), we get

$\frac{n(n-1)}{2}{a}^{2n-2}{b}^2=729\times 30375$ ...(vii)

Squaring both sides of (v), we get

$n^2{a}^{2n-2}{b}^2=(7290{)}^2$ ...(viii)

Dividing (vii) by (viii), we get

$\frac{n(n-1)a^{2n-2}{b}^2}{2n^2{a}^{2n-2}{b}^2}=\frac{729\times 30375}{7290\times 7290}$

$\Rightarrow \frac{(n-1)}{2n}=\frac{30375}{72900}\Rightarrow \frac{n-1}{2n}=\frac{5}{12}$

$\Rightarrow 12n-12=10n$

$\Rightarrow 2n=12\Rightarrow n=6$

From (iv) $a^6=729\Rightarrow a^6=(3{)}^6\Rightarrow a=3$

From (v) $6\times 3^5\times b=7290\Rightarrow b=5$

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