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How do you find the sum of #Sigma [(i-1)^2+(i+1)^3]# where i is [1,4]?

Answer 1

Scarlett Allison

#238#

#sum_(i=1)^4 [(i-1)^2+(i+1)^3]#

#=sum_(i=1)^4 (i^2-2i+1+i^3+3i^2+3i+1)#

#=sum_(i=1)^4 (i^3+4i^2+i+2)#

#=((4*5)/2)^2+4*(4*5*9)/6+(4*5)/2+2*4#

#=238#

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Answer 2

Charles Arocha

To find the sum of the given series ( \sum_{i=1}^{4} [(i-1)^2+(i+1)^3] ), first substitute ( i ) with each value from 1 to 4, then calculate the expression for each value of ( i ), and finally sum up all the results.

( \sum_{i=1}^{4} [(i-1)^2+(i+1)^3] = (1-1)^2 + (1+1)^3 + (2-1)^2 + (2+1)^3 + (3-1)^2 + (3+1)^3 + (4-1)^2 + (4+1)^3 )

( = (0)^2 + (2)^3 + (1)^2 + (3)^3 + (1)^2 + (4)^3 + (3)^2 + (5)^3 )

( = 0 + 8 + 1 + 27 + 1 + 64 + 9 + 125 )

( = 0 + 8 + 1 + 27 + 1 + 64 + 9 + 125 = 235 )

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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.