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Evaluate \[\sum_{k=1}^{7}2^{k+3}\]
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or you can do manually.
can i get it manually lol
yes, pretty easily.
2^4+2^5+2^6+2^7+2^8+2^9+2^10
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16+32+64+128+256+512+1024
Or you can do it the tricky, backwards way. 2^11-2^4 is the same as what rosho posted. That's just 2048-16, and it saves a lot of addition.
The way I do the previous problems is split the sigma into two parts and add them
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