Question

The first three terms in the binomial expansion of $(a+b{)}^n$ are given to be $729,7209,$ and $30375$ respectively. Find $a,b,$ and $n.$

• A. $a=3,b=6,n=5$
• B. $a=6,b=5,n=3$
• C. $a=3,b=5,n=6$
• D. $a=5,b=3,n=6$

Answer 1

Answer: (C)

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

Here, first 3 terms are $a^n{,}^n{C}_1{a}^{n-1}b$ and ${}^n{C}_2{a}^{n-2}{b}^2$.

Also, it is given that their values are $729,7290$ and $30375$.

Therefore,

$a^n=729$ …… (1)

${}^n{C}_1{a}^{n-1}b=7290$ …… (2)

${}^n{C}_2{a}^{n-2}{b}^2=30375$ …… (3)

Divide (1) by (2).

$\frac{a^n}{{}^n{C}_1{a}^{n-1}b}=\frac{729}{7290}$

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

$\frac{a}{nb}=\frac{1}{10}$ …… (4)

Divide (2) by (3).

$\frac{{}^n{C}_1{a}^{n-1}b}{{}^n{C}_2{a}^{n-2}{b}^2}=\frac{7290}{30375}$

$\frac{2}{(n-1)}\times \frac{a^{n-1-(n-2)}}{1}\times \frac{b}{b^2}=\frac{6}{25}$

$\frac{2a}{(n-1)b}=\frac{6}{25}$ …… (5)

Now, divide (4) by (5).

$\frac{\frac{a}{nb}}{\frac{2a}{(n-1)b}}=\frac{\left(\frac{1}{10}\right)}{\left(\frac{6}{25}\right)}$

$\frac{n-1}{2n}=\frac{5}{12}$

$12n-12=10n$

$n=6$

Therefore,

$a^6=729$

$a=3$

And,

$\frac{2a}{(n-1)b}=\frac{6}{25}$

$5b=25$

$b=5$

Hence,

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