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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{1}{(1 - b) (1 - a b)}$
$S = 1 + (1 + a) b + (1 + a + a^{2}) b^{2} + . . . . . \infty$
$S (b) = b + (1 + a) b^{2} . . . . \infty$
$S (1 - b) = 1 + a b + a^{2} b^{2} . . . . . \infty$
$S (1 - b) = \frac{1}{1 - a b}$
$S = \frac{1}{(1 - b) (1 - a b)}$

Suggest Corrections

Similar questions

| a | < 1

| b | < 1

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

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

| a | < 1, | b | < 1

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

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

| a | < 1

| b | < 1

s = 1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + . . . i s - .

Join BYJU'S Learning Program