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write the following in summation notation: 1/2+2/4+3/8+4/16+5/32+6/64
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\[\sum_{k=1}^{6}k/2^k\]
check it, see if it works..
1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64, yeah it does..
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\[\sum_{i-3}^{7} 4i ^{2}\] also do you know how to evaluate this??
4(3)^2+4(4)^2+4(5)^2 ... until 4(7)^2
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