If \(x\) = 1 + a + a2 + ..... ∞ and y = 1 + b + b2 + ...... ∞, whe

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

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»

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

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