Sum of 2^N Sequence from 1-50: Solving Sigma Notation Problem

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SUMMARY

The discussion focuses on calculating the sum of the sequence defined by the sigma notation S = Σ(2^k) from k=1 to k=50. The user proposes a method involving multiplying the sum by 2, resulting in the equation 2S = Σ(2^(k+1)) from k=1 to k=50, which allows for the subtraction of the original sum from the modified sum. This approach leads to a simplified equation that can be solved to find the total sum of the series.

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Okay, I think this problem relates with sigma notation (I'm not good at it)

the sequence given is 2, 4, 8, 16, 32 ...

It wants you to find the sum from 1-50

So the equation thing is 2^N?

How do you find all the sums again...? Thanks
 
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Write out the sum:

S = 2 + 2^2 + \cdot \cdot \cdot + 2^{50}

If you multiply the sum by 2 you get

2S = 2^2 + 2^3 + \cdot \cdot \cdot + 2^{50} + 2^{51}

What happens when you subtract the first equation from the second? Can you determine from that what the sum is?
 
If you mean something like

S=\sum_{k=1}^{50} 2^{k} =...?

, i guess the notation shouldn't be too difficult to grasp.
I hope.

Daniel.
 

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