If x - 1 - a - a2 - a3 - - and y - 1 - b - b2 - b3 - - then show that 1 - ab - a2b2 - a3b3 - - - xyx - y - 1

Given, x = 1 + a + a^2 + a^3 + ... S_∞ = a1 r Here, a = 1, r = a Therefore, x = 11 a ⇒ 1 a = 1x ⇒ a = 1 1x ⇒ y = 1 + ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

If x = 1 + ...

Question

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

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Solution

Given, $x = 1 + a + a^{2} + a^{3} + . . .$
$S_{\infty} = \frac{a}{1 - r}$
Here, $a = 1, r = a$
Therefore, $x = \frac{1}{1 - a}$
$\Rightarrow 1 - a = \frac{1}{x}$
$\Rightarrow a = 1 - \frac{1}{x}$
$\Rightarrow y = 1 + b + b^{2} + b^{3} + . . . .$
$\Rightarrow a = 1, r = b$
$\Rightarrow y = \frac{1}{1 - b}$
$\Rightarrow 1 - b = \frac{1}{y}$
$\Rightarrow b = 1 - \frac{1}{y}$
Taking L.H.S
$1 + a b + a^{2} b^{2} + a^{3} b^{2} + . . . . .$
$= 1 + a b + (a b)^{2} + (a b)^{2} + . . . . .$
$= 1 + a b + (a b)^{2} + (a b)^{3} + . . . . .$
$S_{\infty} = \frac{a}{1 - r}$ Here $a = 1, r = a b$
$= \frac{1}{1 - a b}$
$= \frac{1}{1 - (1 - \frac{1}{x}) (1 - \frac{1}{y})}$
$= \frac{1}{1 - (1 - \frac{1}{y} - \frac{1}{x} + \frac{1}{x y})}$
$= \frac{1}{1 - 1 + \frac{1}{y} + \frac{1}{x} - \frac{1}{x y}}$
$= \frac{1}{\frac{1}{y} + \frac{1}{x} - \frac{1}{x y}} = \frac{1}{\frac{x + y - 1}{x y}}$
$= \frac{x y}{x + y - 1}$

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If x=1+a+a2+a3+... and y=1+b+b2+b3+...., then show that 1+ab+a2b2+a3b3+...=xyx+y−1

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