If a- x are real numbers and - a -1- x -1- then 1-1- a x -1- a - a 2 x 2- isA- 1-1-a1-xB- 1-x1-a xC- -1-a x1-aD- 1-1-a1-a x

The correct option is B 1(1 x)(1 ax)Let S=1+(1+a)x+(1+a+a^2)x^2+.....∞...(1) xS= x+(1+a)x^2+..........∞...(2) Substracting (2) from (1), we get; ...

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Standard XII

Mathematics

Binomial Theorem

If a, x are r...

Question

If $a, x$ are real numbers and $| a | < 1, | x | < 1,$ then $1 + (1 + a) x + (1 + a + a^{2}) x^{2} . . . . . . \infty$ is

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

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

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

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

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Solution

The correct option is C $\frac{1}{(1 - x) (1 - a x)}$
Let $S = 1 + (1 + a) x + (1 + a + a^{2}) x^{2} + . . . . . \infty . . . (1)$
$x S = x + (1 + a) x^{2} + . . . . . . . . . . \infty . . . (2)$
Substracting $(2)$ from $(1)$ , we get;
$(1 - x) S = 1 + a x + a^{2} x^{2} + . . . \infty$
$\Rightarrow (1 - x) S = \frac{1}{1 - a x}$
$\Rightarrow S = \frac{1}{(1 - x) (1 - a x)}$

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If a,x are real numbers and |a|<1,|x|<1, then 1+(1+a)x+(1+a+a2)x2......∞ is

The correct option is C 1(1−x)(1−ax)Let S=1+(1+a)x+(1+a+a2)x2+.....∞...(1) xS= x+(1+a)x2+..........∞...(2) Substracting (2) from (1), we get; (1−x)S=1+ax+a2x2+...∞ ⇒(1−x)S=11−ax ⇒S=1(1−x)(1−ax)

If $a, x$ are real numbers and $| a | < 1, | x | < 1,$ then $1 + (1 + a) x + (1 + a + a^{2}) x^{2} . . . . . . \infty$ is