How do you use the properties of summation to evaluate the sum of #Sigma (i^2+3)# from i=1 to 20?

Answer 1

# sum_(i=1)^20(i^2+3) = 2930 #

You can use the standard result

# sum_(i=1)^ni^2=1/6n(n+1)(2n+1) #

So:

# sum_(i=1)^20(i^2+3) = sum_(i=1)^20(i^2)+ sum_(i=1)^20(3) # # :. sum_(i=1)^20(i^2+3) = 1/6(20)(21)(41)+(20)(3) # # :. sum_(i=1)^20(i^2+3) = 2870+60 # # :. sum_(i=1)^20(i^2+3) = 2930 #

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Answer 2

Aurora Alvarado

You can use the properties of summation to evaluate the sum (\sum_{i=1}^{20} (i^2 + 3)) by splitting it into two separate sums: (\sum_{i=1}^{20} i^2) and (\sum_{i=1}^{20} 3). Then, you can apply the formulas for the sum of squares and the sum of a constant sequence to compute the values.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.