3,−6,12,−24,...
You can rewrite this as:
(−1)n3⋅(1,2,4,8,...)
If we focus on 1,2,4,8...
an+1=2anwith a0=1.
This can be written out as a nice sum:N∑n=02n=20+21+22+23+...=1+2+4+8+...
Thus, now we can recombine everything to get:
3N∑n=0(−1)n2n
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