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Evaluate the summation below. sum_(n=0)^(5)(-2n^(2)-n) Answer:

Evaluate the summation below.\newline∑n=05(−2n2−n) \sum_{n=0}^{5}\left(-2 n^{2}-n\right) \newlineAnswer:

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Q. Evaluate the summation below.\newline∑n=05(−2n2−n) \sum_{n=0}^{5}\left(-2 n^{2}-n\right) \newlineAnswer:
  1. Calculate for n=0n=0: For n=0n=0: (−2(0)2−0)=0(-2(0)^2 - 0) = 0
  2. Calculate for n=1n=1: For n=1n=1: (−2(1)2−1)=−2−1=−3(-2(1)^2 - 1) = -2 - 1 = -3
  3. Calculate for n=2n=2: For n=2n=2: (−2(2)2−2)=−2(4)−2=−8−2=−10(-2(2)^2 - 2) = -2(4) - 2 = -8 - 2 = -10
  4. Calculate for n=3n=3: For n=3n=3: (−2(3)2−3)=−2(9)−3=−18−3=−21(-2(3)^2 - 3) = -2(9) - 3 = -18 - 3 = -21
  5. Calculate for n=4n=4: For n=4n=4: (−2(4)2−4)=−2(16)−4=−32−4=−36(-2(4)^2 - 4) = -2(16) - 4 = -32 - 4 = -36
  6. Calculate for n=5n=5: For n=5n=5: (−2(5)2−5)=−2(25)−5=−50−5=−55(-2(5)^2 - 5) = -2(25) - 5 = -50 - 5 = -55
  7. Sum of all values: Now, we sum all the values obtained for each nn:0+(−3)+(−10)+(−21)+(−36)+(−55)=−1250 + (-3) + (-10) + (-21) + (-36) + (-55) = -125

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