Evaluate sum please help
posted by Crystal
Evaluate the sum
∑3 =0 (1)k × 2k1/2k1
6

Steve
surely (2k1)/(2k1) = 1
So, the sum is
6
∑ (1)^k = (1)^3 + (1)^4 + (1)^5 + (1)^6 = 1+11+1 = 0
3
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