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