Question

If the expansion in power of x of the function $\dfrac{1}{\left( 1-ax \right)\left( 1-bx \right)}$ is ${{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....$ then ${{a}_{n}}$ is
A. $\dfrac{{{b}^{n}}-{{a}^{n}}}{b-a}$
B. $-\dfrac{{{b}^{n}}-{{a}^{n}}}{b-a}$
C. $-\dfrac{{{b}^{n+1}}-{{a}^{n+1}}}{b-a}$
D. $\dfrac{{{b}^{n+1}}-{{a}^{n+1}}}{b-a}$

Answer 1

Find the individual expansion of $\dfrac{1}{\left( 1-ax \right)}$ and $\dfrac{1}{\left( 1-bx \right)}$. Multiply them and form a new GP series. Find the first term, common ratio and put the values in the formula to find the sum of n terms. Simplify and get ${{a}_{n}}$.

Complete step-by-step answer:
We have been given the expression, $\dfrac{1}{\left( 1-ax \right)\left( 1-bx \right)}$.
Thus, we can write it as a negative power series $={{\left( 1-ax \right)}^{-1}}{{\left( 1-bx \right)}^{-1}}$.
Therefore, it becomes an infinite series as given in the question.
We know that, $\dfrac{1}{1-ax}=\sum\limits_{r=0}^{\infty }{{{\left( ax \right)}^{r}}}$.
We have been given the series, ${{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....$
Thus, we need to get the value of ${{a}_{n}}$.
The expansion of ${{\left( 1-ax \right)}^{-1}}=1+ax+{{a}^{2}}{{x}^{2}}+.....$
The expansion of ${{\left( 1-bx \right)}^{-1}}=1+bx+{{b}^{2}}{{x}^{2}}+.....$
The expansion is in GP i.e. Geometric progression. It is a sequence in which each term is derived by multiplying / dividing the preceding term by a fixed number called common ratio, which is denoted by r.
Thus, ${{\left( 1-ax \right)}^{-1}}{{\left( 1-bx \right)}^{-1}}=\left[ 1+ax+{{a}^{2}}{{x}^{2}}+..... \right]\left[ 1+bx+{{b}^{2}}{{x}^{2}}+.....{{b}^{n}}{{x}^{n}} \right]$
In the multiplication we need to find ${{a}_{n}}$ which is coefficient of ${{x}^{n}}$.
So, in both the expansion we can count till the ${{n}^{th}}$ power of x.
Thus, the final term of expansion of ${{\left( 1-bx \right)}^{-1}}$ will be ${{b}^{n}}{{x}^{n}}$.
So, we break ${{\left( 1-ax \right)}^{-1}}{{\left( 1-bx \right)}^{-1}}=\left[ 1+ax+{{a}^{2}}{{x}^{2}}+..... \right]\left[ 1+bx+{{b}^{2}}{{x}^{2}}+.....{{b}^{n}}{{x}^{n}} \right]$ like $\left[ 1+ax+{{a}^{2}}{{x}^{2}}+.....+{{a}^{n-2}}{{x}^{n-2}}+{{a}^{n-1}}{{x}^{n-1}}+{{a}^{n}}{{x}^{n}}+.... \right]\left[ 1+bx+{{b}^{2}}{{x}^{2}}+.....+{{b}^{n-2}}{{x}^{n-2}}+{{b}^{n-1}}{{x}^{n-1}}+{{b}^{n}}{{x}^{n}}+.... \right]$ Thus, multiply the above expression to get coefficient of ${{x}^{n}}$, we get the form of $\left( 1\times {{b}^{n}} \right)+\left( a\times {{b}^{n-1}} \right)+\left( {{a}^{2}}\times {{b}^{n-2}} \right)+...={{b}^{n}}+a{{b}^{n-1}}+{{a}^{2}}{{b}^{n-2}}+.......+{{a}^{n}}$
Thus, the above series becomes GP and now we need to find the common ratio, b is common in all terms. Similarly, a is common in all terms except the first term.
$\therefore$ Common ratio $=\dfrac{a{{b}^{n-1}}}{{{b}^{n}}}=\dfrac{a{{b}^{n}}.{{b}^{-1}}}{{{b}^{n}}}=a{{b}^{-1}}=\dfrac{a}{b}$
Thus, we got the common ratio as $\dfrac{a}{b}$.
The first term of series = ${{b}^{n}}$.
$\because$ By law of exponent, ${{x}^{m-n}}=\dfrac{{{x}^{m}}}{{{x}^{n}}}$. Similarly, ${{b}^{n-1}}=\dfrac{{{b}^{n}}}{b}$.
The sum of n terms of a GP is given by the formula,
${{S}_{n}}=\dfrac{a\left[ {{r}^{n}}-1 \right]}{r-1}$, Put $a={{b}_{n}}$, $r=\dfrac{a}{b},n=n+1$.
Here number of terms, n = n + 1.
$\therefore S=\dfrac{{{b}^{n}}\left[ {{\left( \dfrac{a}{b} \right)}^{n+1}}-1 \right]}{\dfrac{a}{b}-1}=\dfrac{{{b}^{n}}\left[ \dfrac{{{a}^{n+1}}}{{{b}^{n+1}}}-1 \right]}{\dfrac{a-b}{b}}=\dfrac{{{b}^{n+1}}\left[ \dfrac{{{a}^{n+1}}-{{b}^{n+1}}}{{{b}^{n+1}}} \right]}{a-b}$
Canceling out ${{b}^{n+1}}$ from the numerator and simplify the expression as $\dfrac{{{a}^{n+1}}-{{b}^{n+1}}}{a-b}$.
Let us multiply by (-1) in numerator and denominator.
$\dfrac{-\left( {{a}^{n+1}}-{{b}^{n+1}} \right)}{-\left( a-b \right)}=\dfrac{+\left( {{b}^{n+1}}-{{a}^{n+1}} \right)}{+\left( b-a \right)}$
Thus, we got, ${{a}_{n}}=\dfrac{{{b}^{n+1}}-{{a}^{n+1}}}{b-a}$.

So, the correct answer is “Option d”.

Note: We have used Geometric progression here, which is a very important concept in sequence and series. By looking at the series you should be able to determine if it’s an AP, GP or HP in order to solve it. So, it’s important to learn and remember various concepts and formulas in the series.