Question

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

Answer 1

Hint: We know that ${{\left( r+1 \right)}^{th}}$ term in the expansion of ${{\left( a+b \right)}^{n}}$ is given by $^{n+1}{{C}_{r}}{{a}^{n}}{{b}^{r}}$ . hence we will write the first three terms and since we know if the first three terms of the expression are 729, 7290 and 30375 respectively, we will get three equations. Now we will solve the equations by dividing and hence find the value of a, b and n.

Complete step-by-step answer:
Now we know that the binomial expansion of ${{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{b}}{{b}^{0}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{{.....}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}$
Hence the ${{\left( r+1 \right)}^{th}}$ term is given by $^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$ .
Now the first three terms are given to be 729, 7290 and 30375 respectively.
Hence we get
$\begin{align}
  & ^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}=729 \\
 & \Rightarrow {{a}^{n}}=729..............(1) \\
\end{align}$
Now
$\begin{align}
  & ^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}=7290 \\
 & {{\Rightarrow }^{n}}{{C}_{1}}{{a}^{n-1}}b=7290..............\left( 2 \right) \\
\end{align}$
And also
$^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}=30375..............\left( 3 \right)$
Now dividing equation (1) by equation (2)
\[\dfrac{{{a}^{n}}}{^{n}{{C}_{1}}{{a}^{n-1}}b}=\dfrac{729}{7290}\]
We know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ hence we get
\[\begin{align}
  & \dfrac{{{a}^{n}}}{\dfrac{n!}{1!\left( n-1 \right)!}{{a}^{n-1}}b}=\dfrac{1}{10} \\
 & \dfrac{{{a}^{n}}}{\dfrac{n\left( n-1 \right)!}{\left( n-1 \right)!}{{a}^{n-1}}b}=\dfrac{1}{10} \\
 & \dfrac{{{a}^{n}}}{n{{a}^{n-1}}b}=\dfrac{1}{10} \\
\end{align}\]
Now we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ , hence we get.
\[\begin{align}
  & \\
 & \dfrac{{{a}^{n-(n-1)}}}{nb}=\dfrac{1}{10} \\
 & \dfrac{a}{nb}=\dfrac{1}{10}.......................(4) \\
\end{align}\]
Now dividing equation (2) by equation (3) we get
$\dfrac{^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}}{^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}}=\dfrac{7290}{3075}$
Ow again using $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ hence we get
$\begin{align}
  & \dfrac{\dfrac{n!}{\left( n-1 \right)!1!}{{a}^{n-1}}{{b}^{1}}}{\dfrac{n!}{\left( n-2 \right)!2!}{{a}^{n-2}}{{b}^{2}}}=\dfrac{7290}{30375} \\
 & \Rightarrow \dfrac{\dfrac{n\left( n-1 \right)!}{\left( n-1 \right)!1!}{{a}^{n-1}}{{b}^{1}}}{\dfrac{n\left( n-1 \right)\left( n-2 \right)!}{\left( n-2 \right)!2!}{{a}^{n-2}}{{b}^{2}}}=\dfrac{7290}{30375} \\
\end{align}$
Now we know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ hence we get
$\begin{align}
  & \Rightarrow \dfrac{n{{a}^{n-1-\left( n-2 \right)}}{{b}^{1-2}}}{\dfrac{n\left( n-1 \right)}{2!}}=\dfrac{7290}{30375} \\
 & \Rightarrow \dfrac{a{{b}^{-1}}}{\dfrac{\left( n-1 \right)}{2}}=\dfrac{6}{25} \\
\end{align}$
Now multiplying the numerator and denominator of LHS by 2 we get
$\dfrac{2a{{b}^{-1}}}{n-1}=\dfrac{6}{25}$
\[\Rightarrow \dfrac{a{{b}^{-1}}}{n-1}=\dfrac{3}{25}.....................(5)\]
Now dividing equation (5) by equation (4) we get.
\[\begin{align}
  & \dfrac{a{{b}^{-1}}}{n-1}\div \dfrac{a}{nb}=\dfrac{3}{25}\div \dfrac{1}{10} \\
 & \Rightarrow \dfrac{a{{b}^{-1}}}{n-1}\times \dfrac{nb}{a}=\dfrac{3}{25}\times \dfrac{10}{1} \\
 & \Rightarrow \dfrac{n}{n-1}=\dfrac{6}{5} \\
\end{align}\]
Cross multiplying the above equation we get
$5n=6n-6$
Rearranging the terms we get
$6n-5n=6$
Hence we have $n=6$
Now substituting n = 6 in equation (1) we get
$\begin{align}
  & {{a}^{6}}=729 \\
 & \Rightarrow a=\sqrt[6]{729}=\sqrt[6]{3\times 3\times 3\times 3\times 3\times 3} \\
 & \Rightarrow a=3 \\
\end{align}$
Hence we have a = 3 and n = 6.
Now substituting these values in equation (4) we get
$\dfrac{3}{6b}=\dfrac{1}{10}$
Cross multiplying the above equation we get
$30=6b$
Dividing throughout by 6 we get the value of b = 5.
Hence we have a = 3, b = 5, n = 6.

Note: Note that while calculating ${{\left( r+1 \right)}^{th}}$ term of expansion of ${{\left( a+b \right)}^{n}}$ we say that ${{\left( r+1 \right)}^{th}}$ term is $^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$ hence note that if we want to calculate ${{r}^{th}}$ term we will have to substitute $r-1$ in formula $^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$ and hence the formula will be $^{n}{{C}_{r-1}}{{a}^{n-\left( r-1 \right)}}{{b}^{r-1}}$