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How do you find the sum of the finite geometric sequence of #Sigma 5(-3/2)^(n-1)# from n=1 to 8?

Answer 1

Cameron Alexander

#-6305/128#

In general for finite geometric series:

#sum_(k=1)^(n)(a*r^(k-1)) = (a*(1-r^(n)))/(1-r)#

In this case #a=5#, #r=-3/2#, and #n=8#, so:

#sum_(k=1)^(n)(a*r^(k-1)) = (5*(1-(-3/2)^(8)))/(1-(-3/2))=-6305/128#

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Answer 2

Samantha Adkison

To find the sum of the finite geometric sequence represented by Σ 5(-3/2)^(n-1) from n=1 to 8, you can use the formula for the sum of a finite geometric series:

Sum = a(1 - r^n) / (1 - r)

where: a = the first term of the sequence r = the common ratio n = the number of terms in the sequence

In this sequence: a = 5 r = -3/2 n = 8

Plugging these values into the formula:

Sum = 5(1 - (-3/2)^8) / (1 - (-3/2)) Sum ≈ 5(1 - 6561/256) / (1 + 3/2) Sum ≈ 5(1 - 6561/256) / (5/2) Sum ≈ 5(1 - 6561/256) * (2/5) Sum ≈ 2(1 - 6561/256) Sum ≈ 2(1 - 25.664) Sum ≈ 2(0.336) Sum ≈ 0.672

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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.