∑(k=1 to 7) [3 - k2]
= (3-12) + (3-22) + (3-32) + (3-42) + (3-52) + (3-62) + (3-72)
= 2 + (-1) + (-7) + (-13) + (-22) + (-33) + (-46) = -120
Rahul P.
asked • 12/02/18Riemann sums for
7∑ (3- k∧2)
k=1
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