I'm having trouble with a geometric series problem.
Determine if
the infinite summation of
(-3)^(n-1)/4^n
converges or diverges. If it converges, find the sum.
So the answer says that
sigma -3^(n-1)/4^n
= 1/4 * sigma (-3/4)^(n-1)
How did they factor out the 1/4? My algebra is very weak :(
the first part:
4^n can be written as
4(4)^(n-1) = (1/4) (4)^(n-1)
then (-3)^(n-1)/4^n
= (1/4) (-3)^(n-1) / 4^(n-1)
= (1/4) (-3/4)^(n-1)
no now:
∑ (1/4) (-3/4)^(n-1)
= (1/4)(-3/4)^0 + (1/4)(-3/4)^1 + (1/4)(-3/4)^2 + ...
which is an infinitite geometric series , which clearly converges.
with a = (1/4) and r = -3/4
sum∞ = a/(1-r) = (/4)/(1-(-3/4)
= (1/4)/(7/4)
= 1/7
To factor out the 1/4, you can use the property of geometric series. In a geometric series, where each term is multiplied by a constant ratio from the previous term, you can factor out the common ratio.
Let's rewrite the series without the exponentials:
(-3)^(n-1) / 4^n
Now, we can rewrite (-3)^(n-1) as (-3/4) * (-3/4) * ... * (-3/4) (n-1 times). This is based on the idea that (-3/4) raised to the power (n-1) is equal to (-3/4) multiplied by itself (n-1) times.
So, our series becomes:
((-3/4) * (-3/4) * ... * (-3/4)) / 4^n
Now, notice that the numerator (-3/4) repeated (n-1) times is common in each term. Therefore, we can factor it out:
((-3/4)^(n-1) / 4^n) = (1/4) * (-3/4)^(n-1)
And this is how we factor out the 1/4 from the original series.
Now, we have a series in the form:
(1/4) * sigma ((-3/4)^(n-1))
We can work with the factored series to determine if it converges or diverges, and find the sum if it converges.