If x-1-a-a2- and y-1-b-b2- then the value of 1-a b-a2 b2- - - is0- a -1-0- b -1A- x y-1-x-yB- 1-x- y-x-y-1C- x y-2 x y-x-y-1D- x y-x-y-1

The correct option is D xyx+y 1 nbsp;x=1+a+a^2+....∞= 11 a ⇒ x ax=1 ⇒ x 1=ax ⇒ a=x 1x Similarly, b=y 1y Now, 1+ab++a^2b^2+....∞ = nbsp;11 ab =11 (x 1x)× ...

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Standard XI

Mathematics

Relation between Continuity and Differentiability

If x=1+a+a2+…...

Question

If $x = 1 + a + a^{2} + . . . . \infty$ and $y = 1 + b + b^{2} + . . . . \infty,$ then the value of $1 + a b + + a^{2} b^{2} + . . . . \infty$ is
$(0 < a < 1, 0 < b < 1)$

\frac{x y}{1 - x - y}

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\frac{x y}{x + y + 1}

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\frac{x y}{2 x y + x + y - 1}

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\frac{x y}{x + y - 1}

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Solution

The correct option is D $\frac{x y}{x + y - 1}$
$x = 1 + a + a^{2} + . . . . \infty = \frac{1}{1 - a} \Rightarrow x - a x = 1 \Rightarrow x - 1 = a x \Rightarrow a = \frac{x - 1}{x}$
Similarly, $b = \frac{y - 1}{y}$

Now, $1 + a b + + a^{2} b^{2} + . . . . \infty$
$= \frac{1}{1 - a b} = \frac{1}{1 - (\frac{x - 1}{x}) \times (\frac{y - 1}{y})} = \frac{x y}{x y - (x - 1) (y - 1)} = \frac{x y}{x + y - 1}$

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Similar questions

$x = 1 + a + a^{2} + . . . \infty (a < 1)$
$y = 1 + b + b^{2} + . . . \infty (b < 1)$
Then the value of $1 + a b + a^{2} b^{2} + . . . . . \infty$ is
[MNR 1980; MP PET 1985]

If $x = 1 + a + a^{2} . . . . . . . . . . t o \infty$ $(| a | < 1), y = 1 + b + b^{2}$ ......... to $\infty$ (|b| < 1), then $Z = 1 + a b + a^{2}$ $b^{2}$ + $a^{3}$ $b^{3} . . . . . t o \infty$ to is

If x = 1 + a + $a^{2}$ .......... to $\infty$ (|a|<1), y = 1 + b + $b^{2}$ ......... to $\infty$ (|b| < 1),

then Z = 1 + ab + $a^{2}$ $b^{2}$ + $a^{3}$ $b^{3}$ ..... to $\infty$ is

Q. If

x = 1 + a + a^{2} + a^{3} + . . . . . \infty

and

y = 1 + b + b^{2} + b^{3} + . . . . . \infty

; prove that

1 + a b + a^{2} b^{2} + . . . . \infty = \frac{x y}{x + y - 1}

Q. 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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Relation between Continuity and Differentiability

Standard XI Mathematics

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If x=1+a+a2+....∞ and y=1+b+b2+....∞, then the value of 1+ab++a2b2+....∞ is (0<a<1, 0<b<1)

The correct option is D xyx+y−1 x=1+a+a2+....∞=11−a⇒x−ax=1⇒x−1=ax⇒a=x−1x Similarly, b=y−1y Now, 1+ab++a2b2+....∞ =11−ab=11−(x−1x)×(y−1y)=xyxy−(x−1)(y−1)=xyx+y−1

If $x = 1 + a + a^{2} + . . . . \infty$ and $y = 1 + b + b^{2} + . . . . \infty,$ then the value of $1 + a b + + a^{2} b^{2} + . . . . \infty$ is
$(0 < a < 1, 0 < b < 1)$