How do you use the properties of summation to evaluate the sum of #Sigma (i1)^2# from i=1 to 20?
Apply summation formulas to get
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To evaluate the sum of ( \sum_{i=1}^{20} (i1)^2 ), we can first expand the expression ( (i1)^2 ) as ( i^2  2i + 1 ). Then, we can use the properties of summation to separate the sum into three parts: ( \sum_{i=1}^{20} i^2 ), ( 2 \sum_{i=1}^{20} i ), and ( \sum_{i=1}^{20} 1 ).
The formulas for each of these summations are:
 ( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} )
 ( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} )
 ( \sum_{i=1}^{n} 1 = n )
Plugging in ( n = 20 ) into these formulas, we get:
 ( \sum_{i=1}^{20} i^2 = \frac{20(20+1)(220+1)}{6} = \frac{2021*41}{6} = 2870 )
 ( \sum_{i=1}^{20} i = \frac{20(20+1)}{2} = \frac{20*21}{2} = 210 )
 ( \sum_{i=1}^{20} 1 = 20 )
Substituting these values into our original expression, we have: ( \sum_{i=1}^{20} (i1)^2 = \sum_{i=1}^{20} i^2  2 \sum_{i=1}^{20} i + \sum_{i=1}^{20} 1 = 2870  2*210 + 20 = 2870  420 + 20 = 2470 )
Therefore, the sum of ( \sum_{i=1}^{20} (i1)^2 ) is 2470.
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To evaluate the sum ( \sum_{i=1}^{20} (i1)^2 ) using the properties of summation, we first expand the squared term ( (i1)^2 ) and then apply the properties of summation:
[ (i1)^2 = i^2  2i + 1 ]
Now, we can rewrite the sum as follows:
[ \sum_{i=1}^{20} (i^2  2i + 1) ]
Next, we can split this into three separate sums using the properties of summation:
[ \sum_{i=1}^{20} i^2  2 \sum_{i=1}^{20} i + \sum_{i=1}^{20} 1 ]
Now, we evaluate each sum individually:

( \sum_{i=1}^{20} i^2 ): This is a sum of the squares of consecutive integers, which can be evaluated using the formula for the sum of squares: [ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} ] Plugging in ( n = 20 ): [ \sum_{i=1}^{20} i^2 = \frac{20(21)(41)}{6} = 2870 ]

( \sum_{i=1}^{20} i ): This is a sum of consecutive integers, which can be evaluated using the formula for the sum of an arithmetic series: [ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} ] Plugging in ( n = 20 ): [ \sum_{i=1}^{20} i = \frac{20(21)}{2} = 210 ]

( \sum_{i=1}^{20} 1 ): This is simply the sum of the constant term 1, repeated 20 times: [ \sum_{i=1}^{20} 1 = 20 ]
Now, substitute the values back into the original expression:
[ \sum_{i=1}^{20} (i1)^2 = \sum_{i=1}^{20} i^2  2 \sum_{i=1}^{20} i + \sum_{i=1}^{20} 1 = 2870  2 \times 210 + 20 ]
Finally, calculate the result:
[ \sum_{i=1}^{20} (i1)^2 = 2870  420 + 20 = 2470 ]
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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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