$\sum_{k=-200}^{202}2^{2k+4}$

Question

$$\sum_{k=-200}^{202}2^{2k+4}$$ I know I can try putting it in the online calculators and they would give me the answer, but I would like to keep that as my last resort. I am supposed to give an exact answer while showing each step. I have tried putting in $k$ values, for example

$s=2^{-396}$ when $k=-200$. This pattern stays constant until you reach $k=-2$ where $s=1$. Then it's $2^2$, $2^4$, etc.

I do get the pattern, but how can I use what I know to find the exact sum. I need help finding the answer and show each step I take.

Answer 1

Hint: easier task: $$\sum_{k=-2}^{2}2^{2k+4}=2^4\sum_{k=-2}^{2}2^{2k}=2^4\sum_{k=-2}^{2}4^{k}=2^4(4^{-2}+4^{-1}+4^0+4^1+4^2)= \\ 16\cdot \frac{4^{-2}(4^5-1)}{4-1}=341.$$ Now you can use the same method for your sum.

Answer 2

Hints:

1. $$\sum_{k = -x}^{z} a^k \equiv \sum_{k = -x}^{-1} a^k + \sum_{k = 0}^{z} a^k$$
2. $$\sum_{k = -x}^{-y} a^k \equiv \sum_{k = y}^{x} \left(\frac{1}{a}\right)^k$$
3. $$\sum_{k = 0}^{z} a^k$$ is a Geometric Series.
4. $$\sum_{k = 1}^{z} a^k \equiv \left(\sum_{k = 0}^{z} a^k\right) - 1$$

Using these hints, you can easily solve the question.

Answer 3

Hint: If $S$ is the sum you seek, $4S=\sum_{k=-200}^{202}2^{2k +6}=2^{410}+S-2^{-396}$. Can you find $S$?