How do you find the sum of #Sigma 2i^2# where i is [0,5]?
# sum_(i=0)^5 2i^2 = 110 #
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To find the sum of the series represented by the sigma notation ( \sum_{i=0}^{5} 2i^2 ), you follow these steps:

Understand the Sigma Notation: The given notation means that you are summing the values of (2i^2) for every integer (i) starting from 0 up to 5.

Calculate Each Term: Calculate the value of (2i^2) for each (i) in the range [0,5] and then sum these values. The formula (2i^2) means you square (i), and then multiply the result by 2.
 For (i = 0), (2 \times 0^2 = 0)
 For (i = 1), (2 \times 1^2 = 2)
 For (i = 2), (2 \times 2^2 = 8)
 For (i = 3), (2 \times 3^2 = 18)
 For (i = 4), (2 \times 4^2 = 32)
 For (i = 5), (2 \times 5^2 = 50)

Sum the Results: Add all the calculated values together.
Let's compute the total sum.The sum of the series represented by ( \Sigma_{i=0}^{5} 2i^2 ) is 110.
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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