Answer
$-73$
Work Step by Step
$\sum\limits_{k=1}^6(3-k^2)=\sum\limits_{k=1}^63-\sum\limits_{k=1}^6k^2=6\cdot3-(1^2+2^2+3^2+4^2+5^2+6^2)=18-(1+4+9+16+25+36)=18-91=-73$
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