If - a -1 and - b -1- then the sum of the series 1-1-a b-1-a-a2 b2-1-a-a2-a3 b3- isA- 1-a1-bB- 1-b1-a bC- 1-1-a1-bD- 1-1-a1-b1-a b

The correct option is B 1(1 b)(1 ab)1 + (1 + a)b + (1 + a + a^2)b^2 + (1 + a + a^2+a^3)b^3+⋯ nbsp;= ∑ _n=1^∞ (1 +a +a^2 +⋯ +a^n 1)b^n 1 nbsp;= ∑ _n=1^∞ ...

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Byju's Answer

Standard XII

Mathematics

Geometric Progression

If | a |<1 an...

Question

If $| a | < 1$ and $| b | < 1$ , then the sum of the series
$1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + \dots$ is

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

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

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

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

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Solution

The correct option is B $\frac{1}{(1 - b) (1 - a b)}$
$1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + \dots = \infty \sum n = 1 (1 + a + a^{2} + \dots + a^{n - 1}) b^{n - 1} = \infty \sum n = 1 (\frac{1 - a^{n}}{1 - a}) b^{n - 1} = \infty \sum n = 1 \frac{b^{n - 1}}{1 - a} - \infty \sum n = 1 \frac{a^{n} b^{n - 1}}{1 - a} = \frac{1}{1 - a} \infty \sum n = 1 b^{n - 1} - \frac{a}{1 - a} \infty \sum n = 1 (a b)^{n - 1} = \frac{1}{1 - a} \times \frac{1}{1 - b} - \frac{a}{1 - a} \times \frac{1}{1 - a b} = \frac{1}{(1 - b) (1 - a b)}$

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Similar questions

If $| a | < 1 a n d | b | < 1$ , then the sum fo the series $1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + . . . i s$

Q. If

a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} \in R

and are such that

a_{i} b_{j} \neq 1

for

1 \leq i, j \leq 3

, then

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1 - a_{1}^{3} b_{1}^{3}}{1 - a_{1} b_{1}} & \frac{1 - a_{1}^{3} b_{2}^{3}}{1 - a_{1} b_{2}} & \frac{1 - a_{1}^{3} b_{3}^{3}}{1 - a_{1} b_{3}} \frac{1 - a_{2}^{3} b_{1}^{3}}{1 - a_{2} b_{1}} & \frac{1 - a_{2}^{3} b_{2}^{3}}{1 - a_{2} b_{2}} & \frac{1 - a_{2}^{3} b_{3}^{3}}{1 - a_{2} b_{3}} \frac{1 - a_{3}^{3} b_{1}^{3}}{1 - a_{3} b_{1}} & \frac{1 - a_{3}^{3} b_{2}^{3}}{1 - a_{3} b_{2}} & \frac{1 - a_{3}^{3} b_{3}^{3}}{1 - a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

> 0 Provided either

a_{1} < a_{2} < a_{3}

and

b_{1} < b_{2} < b_{3}

, or

a_{1} > a_{2} a_{3}

and

b_{1} > b_{2} > b_{3}

then show

(a_{1} - a_{2}) (a_{2} - a_{3}) (a_{3} - a_{1}) (b_{1} - b_{2}) (b_{2} - b_{3}) (b_{3} - b_{1}) < 0

Q. If

| a | < 1

| b | < 1

, then the sum of the series

1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + . . . .

Q. Two curves

a_{i} x^{2} + b_{i} y^{2} = 1; i = 1, 2

where

a_{1} \neq a_{2}, b_{1} \neq b_{2}, a_{1}, a_{2}, b_{1}, b_{2} \neq 0

may intersect orthogonally if _____

Q. If

| a | < 1, | b | < 1

then the sum of the series

1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} +

………… to infinite terms is

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If |a|<1 and |b|<1, then the sum of the series 1+(1+a)b+(1+a+a2)b2+(1+a+a2+a3)b3+⋯ is

The correct option is B 1(1−b)(1−ab)1+(1+a)b+(1+a+a2)b2+(1+a+a2+a3)b3+⋯=∞∑n=1(1+a+a2+⋯+an−1)bn−1=∞∑n=1(1−an1−a)bn−1=∞∑n=1bn−11−a−∞∑n=1anbn−11−a=11−a∞∑n=1bn−1−a1−a∞∑n=1(ab)n−1=11−a×11−b−a1−a×11−ab=1(1−b)(1−ab)

If $| a | < 1$ and $| b | < 1$ , then the sum of the series
$1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + \dots$ is