Question

If \(x\) = 1 + a + a² + ..... ∞ and y = 1 + b + b² + ...... ∞, where a and b are proper fractions, then 1 + ab + a²b² + ... ∞ equals

(a) \(\frac{x+y}{x-y}\)

(b) \(\frac{x^2+y^2}{x-y}\)

(c) \(\frac{x^2+y^2}{x+y-1}\)

(d) \(\frac{xy}{x+y-1}\)

Answer 1

(d) \(\frac{xy}{x+y-1}\)

Since a and b are proper fractions, | a | < 1, | b | < 1 

∴ \(x\) = 1 + a + a² + ..... ∞ = \(\frac{1}{1-a}\)           \(\big(\because\,S_\infty=\frac{a}{1-r}\big)\)

and y = 1 + b + b² + ..... ∞ = \(\frac{1}{1-b}\) 

Also, 1 + ab + a²b² + .....∞ = \(\frac{1}{1-ab}\)        ....(i)  

Now  \(x\) = \(\frac{1}{1-a}\) ⇒ \(x\) – \(x\)a = 1

⇒ \(x\)a = \(x\) – 1 ⇒ a = \(\frac{x-1}{x}\)                 .....(ii)

y = \(\frac{1}{1-b}\)  ⇒ y – yb = 1

⇒ yb = y – 1 ⇒ b = \(\frac{y-1}{y}\)                  .....(iii)

∴ Putting the values of a & b from (ii) and (iii) in (i), we get 

Reqd. sum = \(\frac{1}{1-ab}\) = \(\frac{1}{1-\big(\frac{x-1}{x}\big)\big(\frac{y-1}{y}\big)}\)

= \(\frac{xy}{xy-(xy-x-y+1)}\) = \(\frac{xy}{x+y-1}\).