Question

a1+a2+a3+ ... is an infinite geometric series whose sum is 3. Replacing each of the terms of the series by their squares results in a series whose sum is the same. Replacing each of the terms of the series by their cubes results in a series whose sum can be expressed by a/b where a and b are co-pime positive integers. What is a+b?​

Answer 1

Answer:

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Answer 2

Replacing each of the terms of the series by their cubes results in a series whose sum can be expressed by 27/7

The sum of 27 and 7 is 34

a + b = 34

Step-by-step explanation:

Let the first term of the GP be a and common ratio r

Then

$a_1=a$

$a_2=ar$

$a_3=ar^2$

We know that sum of infinite GP whose first term is a and common ratio r is

$S=\frac{a}{1-r}$

According to the question,

$\frac{a}{1-r}=3$ ............ (1)

Replacing each term of the series with their squares, we will get GP as

$a^2, a^2r^2, a^2r^4, ....$

The first term of this GP is $a^2$ and common ratio $r^2$

Hence, the sum of this infinite GP

$S'=\frac{a^2}{1-r^2}$

Given, S' = 3

Therefore,

$\frac{a^2}{1-r^2}=3$  ............ (2)

Squaring equation (1) and dividing it by eq (2)

$\frac{a^2}{(1-r)^2}\times \frac{1-r^2}{a^2}=\frac{9}{3}$

$\implies \frac{(1-r)(1+r)}{(1-r)^2}=3$

$\implies \frac{1+r}{1-r}=3$

$\implies 1+r=3-3r$

$\implies r=\frac{1}{2}$

Therefore, from (1)

$\frac{a}{1/2}=3$

$\implies a=\frac{3}{2}$

Replacing each term by cubes we get the GP as

$a^3, a^3r^3, a^3r^6, ....$

The sum of this GP is

[tex[S"=\frac{a^3}{1-r^3}[/tex]

$\implies S"=\frac{(3/2)^3}{1-(1/2)^3}$

$\implies S"=\frac{27/8}{7/8}$

$\implies S"=\frac{27}{7}$

Here, a = 27, b = 7

a + b = 27 + 7 = 34

Hope this answer is helpful.

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