Consider two infinite geometric series. The first has leading term $a,$ common ratio $b,$ and sum $S.$ The second has a leading term $b,$ common ratio $a,$ and sum $1/S.$ Find the value of $a+b.$
How about this:
From Damons:
a/(1-b) = s and
b/(1-a) = 1/s OR (1-a)/b = s
so a/(1-b) = (1-a)/b
ab = 1 - a - b + ab
a+b = 1
well, a+b = s + 1/s - (a/s+ bs)
I think that only works if s = 1
a+b = 2 - (a+b)
2(a+b) =2
a+b =1
a ab ab^2 ab^3 .....
infinite so sum = a/ (1-b) = s
b ba ba^2 ba^3 ....
sum = b/(1-a) = 1/s
a = s -bs
b = 1/s -a/s
a+b = s - b s + 1/s -a/s
as a number?
Thanks bro
To solve this problem, let's break it down step by step.
For the first geometric series, we know that the leading term is $a$ and the common ratio is $b.$ The formula to find the sum of an infinite geometric series is given by:
\[S = \frac{a}{1 - b}.\]
Now let's consider the second geometric series. The leading term is $b,$ and the common ratio is $a.$ We want to find the sum, which is denoted as $1/S.$ So, we can write this as:
\[\frac{1}{S} = \frac{b}{1 - a}.\]
To find the value of $a + b,$ we need to solve these two equations simultaneously. We can start by rearranging the equation for the second series:
\[\frac{1}{S} = \frac{b}{1 - a} \Rightarrow S = \frac{1 - a}{b}.\]
Now, we can equate the expressions for $S$ obtained from both series and solve for $a$ and $b.$
\[\frac{a}{1 - b} = \frac{1 - a}{b}.\]
Cross-multiplying, we get:
\[ab = (1 - a)(1 - b).\]
Expanding the right side, we have:
\[ab = 1 - a - b + ab.\]
Rearranging the terms, we get:
\[a + b = 1.\]
Therefore, the sum of the leading terms $a$ and $b$ is equal to $1.$ Hence, $a+b= \boxed{1}.$