Find the value of $b$ given that \[ \dfrac b 7 + \dfrac {b^2}{7^2} + \dfrac {b^3}{7^3} + \cdots = 7.\]

Guest Mar 11, 2023

#1**0 **

We can use the formula for the sum of an infinite geometric series to solve this problem:

S = a/(1-r)

where S is the sum of the series, a is the first term of the series, and r is the common ratio of the series.

In this case, we have:

a = b/7 r = b/7

Plugging these values into the formula, we get:

7 = (b/7)/(1 + b/7)

Multiplying both sides by (1 + b/7), we get:

7 + 7b/7 = b/7

Simplifying, we get:

7 + b = b/7

Multiplying both sides by 7, we get:

49 + 7b = b

Subtracting 7b from both sides, we get:

49 = -6b

Dividing both sides by -6, we get:

b = -49/6

Therefore, the value of b is -49/6.

Guest Mar 11, 2023

#2**0 **

Sum = F / [1 - R], where =First term, R==constant ratio

7 = (49/8) / [1 - R], solve for R

R= 1 /8

b =49 / 8 = 6+1/8

Guest Mar 11, 2023

#3**0 **

I apologize for my mistake. You are correct that b must be less than 7.

Using the formula for the sum of an infinite geometric series, we have:

S = b/7 + b^2/7^2 + b^3/7^3 + b^4/7^4 + ... + b^n/7^n

This is a geometric series with first term b/7 and common ratio b/7. The sum of an infinite geometric series is:

S = a / (1 - r)

where a is the first term and r is the common ratio.

Substituting a = b/7 and r = b/7, we get:

S = (b/7) / (1 - b/7)

Simplifying the denominator, we get:

S = (b/7) / (7-b)/7

Simplifying further, we get:

S = b / (7-b)

Since the sum of the series is equal to 7, we have:

b / (7-b) = 7

Multiplying both sides by (7-b), we get:

b = 7(7-b)

Expanding, we get:

b = 49 - 7b

Solving for b, we get:

8b = 49

**b = 49/8**

**Since b must be less than 7, the solution is:**

**b = 49/8 = 6.125**

**Therefore, the value of b that satisfies the equation is 6.125.**

Guest Mar 11, 2023