How do you find the sum given #Sigma (i-1)^2+(i+1)^3# from i=1 to 4?

Answer 1

# sum_ (i=1)^4 (i-1)^2+(i+1)^3 = 238 #

Let

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

Expanding we have:

# S = (0^2+2^3) + (1^2+3^3) + (2^2+4^3) + (3^2+5^3) # # \ \ = 8 + (1+27) + (4+64) + (9+125) # # \ \ = 8 + 28 + 68 + 134 # # \ \ = 238 #

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Answer 2

Charles Arocha

To find the sum of the given expression ( \Sigma (i-1)^2 + (i+1)^3 ) from ( i=1 ) to ( i=4 ), we can evaluate the expression for each value of ( i ) within the specified range and then add them together.

When ( i = 1 ): [ (1-1)^2 + (1+1)^3 = 0^2 + 2^3 = 0 + 8 = 8 ]

When ( i = 2 ): [ (2-1)^2 + (2+1)^3 = 1^2 + 3^3 = 1 + 27 = 28 ]

When ( i = 3 ): [ (3-1)^2 + (3+1)^3 = 2^2 + 4^3 = 4 + 64 = 68 ]

When ( i = 4 ): [ (4-1)^2 + (4+1)^3 = 3^2 + 5^3 = 9 + 125 = 134 ]

Finally, we add these results together: [ 8 + 28 + 68 + 134 = 238 ]

So, the sum of the expression from ( i=1 ) to ( i=4 ) is ( 238 ).

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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.