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

Parcly Taxel
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CuriousMind
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  • Hint: geometric series. $2^{2k+4} = 2^{2k}2^4 = 16\times 4^k$. – Rócherz Mar 12 '18 at 04:49
  • @CuriousMind Do you know how to add powers together? – an4s Mar 12 '18 at 05:05
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    If the base of two numbers being multiplied is same, then you can add the powers. For example, $16\times32 = 2^4\times2^5$. Since the base of both numbers is same, i.e. $2$, we can add the powers up i.e. $2^{4 + 5} = 2^9$ i.e. $16\times32\equiv2^9$. This is essentially what Rocherz did in his comment, the only difference being instead of combining a product of multiple numbers with same bases into a single number, he took apart a single number into a product of multiple numbers with same bases. – an4s Mar 12 '18 at 05:12
  • Hint 2: $x^{-y} \equiv \left(\dfrac{1}{x}\right)^y$ – an4s Mar 12 '18 at 05:15
  • Hint 3: $$\sum_{k = -x}^{-y} a^k \equiv \sum_{k = y}^{x} \left(\dfrac{1}{a}\right)^k$$ – an4s Mar 12 '18 at 05:25
  • An exact answer as a decimal would have hundreds of digits. In binary, it would be even longer, but much simpler. –  Mar 12 '18 at 07:05

3 Answers3

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.

an4s
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Hint: If $S$ is the sum you seek, $4S=\sum_{k=-200}^{202}2^{2k +6}=2^{410}+S-2^{-396}$. Can you find $S$?

Macavity
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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.

farruhota
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