Calculation of $ \sum_{n=0}^{\infty}2^{2n} z^{2n} $
The answer is
- We note that the n-th summand has the form $(2z)^n$ Denoting w = 2z The sum is sigma of 0 to n summand being $(w)^n$ which can be transformed into 1/(1-w) Rewriting in closed form is $1/(1-2z)$
I don't understand the statement shouldn't you be transforming $w = z^2$ why do you change the form into $(2z)^n$??
Can please help me out? Thank you, sorry for the style
Either the answer or the question is misstated.
If the question is right, then the series is
$$\sum_{n=0}^{\infty}2^{2n}z^{2n} = 1 + 2^2z^2 + 2^4z^4 + 2^6z^6+\cdots$$ $$= 1+(2^2z^2)^1 +(2^2z^2)^2 + (2^2z^2)^3 + \cdots$$
This is a geometric series with ratio $r=2^2z^2$, so the sum is $$\frac{1}{1-r} = \boxed{\dfrac{1}{1-2^2z^2}}$$
If the question is really supposed to be to evaluate $$\sum_{n=0}^{\infty}2^{n}z^{n}$$ then the series is
$$\sum_{n=0}^{\infty}2^{n}z^{n} = 1 + 2^1z^1 + 2^2z^2 + 2^3z^3+\cdots$$ $$= 1+(2z)^1 +(2z)^2 + (2z)^3 + \cdots$$
which is a geometric series with ratio $r = 2z$, so the sum would be $$\frac{1}{1-r} = \boxed{\dfrac{1}{1-2z}}$$