How do you evaluate the series #Sigma 5r# from r=3 to 8?

Answer 1

#sum_(r=3)^8 5r = 165#

We seek:

# sum_(r=3)^8 5r = 5sum_(r=3)^8 r #

Due to the small number of terms required we can just expand the individual terms to get:

# sum_(r=3)^8 5r = 5(3+4+5+6+7+8} # # " " = 5(33) # # " " = 165 #

If the number of individual terms were larger this would be quite cumbersome and use of the standard summation formula

# sum_(r=1)^n r = 1/2n(n+1) #

would be more appropriate. If we used this approach we would get

# sum_(r=3)^8 5r = 5{(sum_(r=1)^8) r - (sum_(r=1)^2 r )}# # " " = 5{1/2*8*9 - 1/2*2*3} # # " " = 5/2{72 - 6} # # " " = 5/2*66 # # " " = 165 #
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Answer 2

To evaluate the series ( \sum_{r=3}^{8} 5r ), you would sum up the terms of the series by plugging in each value of ( r ) from 3 to 8 into the expression ( 5r ), and then adding up the results.

( \sum_{r=3}^{8} 5r = 5(3) + 5(4) + 5(5) + 5(6) + 5(7) + 5(8) )

( = 15 + 20 + 25 + 30 + 35 + 40 )

( = 165 )

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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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