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I have the following sequence:

enter image description here

I need to find the sum of it. However I know that the sequence is not geometrical or arithmetic, therefore I don't know which formula would I use to find the sum.

I wrote a general sum equation: enter image description here

and I'm trying to tackle the problem this way but this problem does not involve higher mathematics, therefore I know that there is a much more easier solution to it.

As you can see wolfram alpha app for series gave me the correct answer, but I would need the steps if possible.

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    $\begingroup$ Use binomial expansion of $(2-1)^{100}$ $\endgroup$ Jul 24 '16 at 12:41
  • $\begingroup$ I can't believe this is so easy.. tnx @SimpleArt $\endgroup$ Jul 24 '16 at 12:44
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$$\begin{align} & \sum_{n=0}^{100}2^{100-n}(-1)^n\binom{100}{n} \\ & =\binom{100}{0}2^{100}-\binom{100}{1}2^{99}+\binom{100}{2}2^{98}- \ldots +\binom{100}{98}2^{2}-\binom{100}{99}2^{1}+\binom{100}{100}2^{0} \\ & =(1-2)^{100} \\ & =(-1)^{100} \\ & =1 \end{align}$$

Please refer to binomial expansion.

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  • $\begingroup$ I will, txn for the link $\endgroup$ Jul 24 '16 at 12:48
  • $\begingroup$ @eugensunic You're welcome. $\endgroup$ Jul 24 '16 at 12:48
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$$ \forall n \in \mathbb{N}, \qquad \sum_{k=1}^n \dbinom{n}{k} 2^{n-k} (-1)^k = \left(2 - 1\right)^n = (-1)^n.$$

Put $n = 100$ and you're done.

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  • $\begingroup$ You should make that for all n even.... For odd n we get -1 $\endgroup$ Jul 24 '16 at 14:24
  • $\begingroup$ Good point, thank you! I edited. $\endgroup$
    – Matt
    Jul 24 '16 at 14:27

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