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whpalmer4Best ResponseYou've already chosen the best response.1
This one can be solved without any cleverness. \[\sum_{k=1}^{4}(1)^k(k+11) =\]\[(1)^1(1+11)+(1)^2)(2+11)+(1)^3)(3+11)+(1)^4(4+11) =\]
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Now why did I stick in those unwanted )'s? \[(1)^1(1+11)+(1)^2(2+11)+(1)^3(3+11)+(1)^4(4+11) =\] For odd k, \((1)^k = 1\) For even k, \((1)^k = 1\) (1+11)+(2+11)(3+11)+(4+11) = all the alternating of signs cancels out the 11s, so it's just 1 + 2  3 + 4 = 2
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
How did you get the 1+23+4?
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
When I solved it I didnt get none of that, maybe I did it wrong?
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[(1+11)+(2+11)(3+11)+(4+11) = 1 11 + 2 + 11  3  11 + 4 + 11\] The alternating 11 + 11 sum to 0, leaving 1 + 2  3 + 4
 one year ago
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