Question

If $x=1+a+{{a}^{2}}+{{a}^{3}}+...$ and $y=1+b+{{b}^{2}}+{{b}^{3}}+...$, then show that $1+ab+{{a}^{2}}{{b}^{2}}+{{a}^{3}}{{b}^{3}}+...=\dfrac{xy}{x+y-1}$

Answer 1

Hint:By using binomial theorem we can write $x={{\left( 1-a \right)}^{-1}},y={{\left( 1-b \right)}^{-1}}$.Now check if $\dfrac{xy}{x+y-1}$ is ${{\left( 1-ab \right)}^{-1}}$ or not, by substituting the values of x and y.

Complete step by step answer:
The binomial theorem is the method of expanding an expression which has been raised to any finite power.
We know that the general form of binomial expansion for negative integer power is,
${{\left( 1-x \right)}^{-n}}=1+nx+\dfrac{n\left( n+1 \right)}{1\times 2}{{x}^{2}}+\dfrac{n\left( n+1 \right)\left( n+2 \right)}{1\times 2\times 3}{{x}^{3}}+...$
It is an infinite series. For this problem, $n=1$.
Therefore,
${{\left( 1-x \right)}^{-1}}=1+1\times x+\dfrac{1\times \left( 1+1 \right)}{1\times 2}{{x}^{2}}+\dfrac{1\times \left( 1+1 \right)\times \left( 1+2 \right)}{1\times 2\times 3}{{x}^{3}}+...$
$\Rightarrow {{\left( 1-x \right)}^{-1}}=1+x+{{x}^{2}}+{{x}^{3}}+...$
 By using binomial theorem for negative integer power we can write,
$x={{\left( 1-a \right)}^{-1}}\Rightarrow x=\dfrac{1}{1-a}......(1)$
Similarly,
$y={{\left( 1-b \right)}^{-1}}\Rightarrow y=\dfrac{1}{1-b}.......(2)$
Now let us multiply equation (1) by equation (2).
$xy=\dfrac{1}{1-a}\times \dfrac{1}{1-b}=\dfrac{1}{\left( 1-a \right)\left( 1-b \right)}......(3)$
If we add equation (1) and equation (2) we will get,
$x+y=\dfrac{1}{1-a}+\dfrac{1}{1-b}$
$\Rightarrow x+y=\dfrac{1-b+1-a}{\left( 1-a \right)\left( 1-b \right)}$
$\Rightarrow x+y=\dfrac{2-a-b}{\left( 1-a \right)\left( 1-b \right)}......(4)$
Now if we subtract 1 from both the sides of equation (3) we will get,
$x+y-1=\dfrac{2-a-b}{\left( 1-a \right)\left( 1-b \right)}-1$
$\Rightarrow x+y-1=\dfrac{2-a-b-\left( 1-a \right)\left( 1-b \right)}{\left( 1-a \right)\left( 1-b \right)}$
Now put $\left( 1-a \right)\left( 1-b \right)=1-a-b+ab$ in the above expression.
$\Rightarrow x+y-1=\dfrac{2-a-b-\left( 1-a-b+ab \right)}{\left( 1-a \right)\left( 1-b \right)}$
$\Rightarrow x+y-1=\dfrac{2-a-b-1+a+b-ab}{\left( 1-a \right)\left( 1-b \right)}$
We can cancel out a few terms from the numerator. Therefore,
$\Rightarrow x+y-1=\dfrac{1-ab}{\left( 1-a \right)\left( 1-b \right)}......(5)$
Now we will divide equation (3) by equation (5).
$\dfrac{xy}{x+y-1}=\dfrac{\dfrac{1}{\left( 1-a \right)\left( 1-b \right)}}{\dfrac{1-ab}{\left( 1-a \right)\left( 1-b \right)}}$
$\Rightarrow \dfrac{xy}{x+y-1}=\dfrac{1}{\left( 1-a \right)\left( 1-b \right)}\times \dfrac{\left( 1-a \right)\left( 1-b \right)}{1-ab}$
We can cancel out the common terms from the numerator and the denominator.
$\Rightarrow \dfrac{xy}{x+y-1}=\dfrac{1}{1-ab}$
$\Rightarrow \dfrac{xy}{x+y-1}={{\left( 1-ab \right)}^{-1}}$
By using binomial expansion for negative integer power we can write,
$\Rightarrow \dfrac{xy}{x+y-1}=1+ab+{{a}^{2}}{{b}^{2}}+{{a}^{3}}{{b}^{3}}+...$
Therefore,
$1+ab+{{a}^{2}}{{b}^{2}}+{{a}^{3}}{{b}^{3}}+...=\dfrac{xy}{x+y-1}$

Note: In this problem it is very important to use the concept of Binomial Theorem. It will make the solution easier. If we try to solve it directly, that means if we put the values of x and y on the right hand side, then the multiplications will be huge and complicated as both the values of x and y are infinite series.