#1**0 **

The given equation is a geometric series with first term b and common ratio 71. The sum of a geometric series is given by the formula S=a(1−rn)/(1-r), where a1 is the first term, r is the common ratio, and n is the number of terms. In this case, we have S=7, a_1=b, and r=1/7.

Solving for n, we get n=−log(1−r)/log(r)=log(1/7)/log(7)=−log(7)/log(7)=−1. Substituting this value of n into the formula for the sum of a geometric series, we get S=a(1-r^n)/(1-r)=b(1-(1/7)^(-1))/(1-1/7)=b(1−7)/(-6)=7. Solving for b, we get **b=49/6**.

Guest May 26, 2023

#2**0 **

sumfor(b, 1, 1000, (6.125^b / (7^b))==7

**b ==6 + 1/8 ==6.125**

These are the first 10 terms of GS:

(7 / 8, 49 / 64, 343 / 512, 2401 / 4096, 16807 / 32768, 117649 / 262144, 823543 / 2097152, 5764801 / 16777216, 40353607 / 134217728, 282475249 / 1073741824)

Guest May 26, 2023

edited by
Guest
May 26, 2023

#3**0 **

The given equation is a geometric series with first term b and common ratio 1/7. The sum of an infinite geometric series is equal to a/(1-r), where a is the first term and r is the common ratio. In this case, the sum is equal to 7, so we have:

b/7 + b^2/7^2 + b^3/7^3 + ... = 7

\frac{b}{1-\frac{1}{7}} = 7

\frac{b}{6} = 7

b = \(\boxed{42}\)

Guest May 26, 2023