1-1-ax-1-a-a2x2- where 0-a- x-1- is equals to

The correct option is C 1(1 x)(1 ax) Let S = 1+ (1+a)x + (1+a+a^2)x^2 + … xS = x+ (1+a)x^2+(1+a+a^2)x^3+… ∴ (1 x)S=1+ax+a^2x^2+…=11 ax ...

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Standard XII

Mathematics

Binomial Coefficients

1+1+ax+1+a+a2...

Question

$1 + (1 + a) x + (1 + a + a^{2}) x^{2} + \dots;$ where $0 < a, x < 1$ , is equals to

\frac{1}{(1 - x) (1 - a)}

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\frac{1}{(1 - a) (1 - a x)}

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\frac{1}{(1 - x) (1 - a x)}

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\frac{1}{(1 - x) (1 + a)}

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Solution

The correct option is C $\frac{1}{(1 - x) (1 - a x)}$
Let $S = 1 + (1 + a) x + (1 + a + a^{2}) x^{2} + \dots$
$x S = x + (1 + a) x^{2} + (1 + a + a^{2}) x^{3} + \dots$
$∴ (1 - x) S = 1 + a x + a^{2} x^{2} + \dots = \frac{1}{1 - a x}$
$\Rightarrow S = \frac{1}{(1 - x) (1 - a x)}$
Hence, option C is correct.

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Similar questions

Q. If

a, x

are real numbers and

| a | < 1, | x | < 1,

then

1 + (1 + a) x + (1 + a + a^{2}) x^{2} . . . . . . \infty

If $¯ x$ is the mean of $x_{1}, x_{2} \dots, x_{n}$ then for $a \neq 0$ , the mean of $a x_{1}, a x_{2} \dots, a x_{n}, \frac{x_{1}}{a}, \frac{x_{2}}{a}, \dots, \frac{x_{n}}{a}$ is
(a) $(a + \frac{1}{a}) ¯ x$
(a) $(a + \frac{1}{a}) \frac{¯ x}{2}$
(c) $(a + \frac{1}{a}) \frac{¯ x}{n}$
(d) $\frac{(a + \frac{1}{a}) ¯ x}{2 n}$

Q. The sum of

1 + (1 + a) x + (1 + a + a^{2}) x^{2} + . . . . \infty, 0 < a, x < 1

equals

Q. Find the sum of

\frac{1}{(1 + x) (1 + a x)} + \frac{a}{(1 + a x) (1 + a^{2} x)} + \frac{a^{2}}{(1 + a^{2} x) (1 + a^{3} x)} + . . . .

n

terms.

Q. Evaluate

{tan}^{- 1} (\frac{\sqrt{1 + a^{2} x^{2}} - 1}{a x})

, where

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1+(1+a)x+(1+a+a2)x2+…; where 0<a,x<1, is equals to

The correct option is C 1(1−x)(1−ax)Let S=1+(1+a)x+(1+a+a2)x2+…xS=x+(1+a)x2+(1+a+a2)x3+…∴(1−x)S=1+ax+a2x2+⋯=11−ax⇒S=1(1−x)(1−ax)Hence, option C is correct.

$1 + (1 + a) x + (1 + a + a^{2}) x^{2} + \dots;$ where $0 < a, x < 1$ , is equals to