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

Answer 1
#sum_(i=1)^20 (i-1)^2 = sum_(i=1)^20 (i^2-2i+1)#
# = sum_(i=1)^20 i^2+sum_(i=1)^20 (-2i)+sum_(i=1)^20 1#
# = sum_(i=1)^20 i^2 -2sum_(i=1)^20 i+sum_(i=1)^20 1#

Apply summation formulas to get

# = ((20)(21)(41))/6-2 ((20)(21))/2 + 20#
# = ((10)(7)(41))-2 ((10)(21)) + 20#
# = 2870 - 420 + 20#
# = 2470#
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Answer 2

To evaluate the sum of ( \sum_{i=1}^{20} (i-1)^2 ), we can first expand the expression ( (i-1)^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} (i-1)^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} (i-1)^2 ) is 2470.

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Answer 3

To evaluate the sum ( \sum_{i=1}^{20} (i-1)^2 ) using the properties of summation, we first expand the squared term ( (i-1)^2 ) and then apply the properties of summation:

[ (i-1)^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:

  1. ( \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 ]

  2. ( \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 ]

  3. ( \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} (i-1)^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} (i-1)^2 = 2870 - 420 + 20 = 2470 ]

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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.

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.

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.

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.

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