How do you find the 4-th partial sum of the infinite series #sum_(n=1)^oo(3/2)^n# ?

Answer 1

To find the fourth partial sum of the infinite series (\sum_{n=1}^{\infty} (3/2)^n), use the formula for the partial sum of a geometric series:

[S_n = a \frac{1 - r^n}{1 - r}]

Where (a) is the first term of the series, (r) is the common ratio, and (n) is the number of terms. In this series, (a = \frac{3}{2}) and (r = \frac{3}{2}).

[S_4 = \frac{3}{2} \frac{1 - (\frac{3}{2})^4}{1 - \frac{3}{2}}]

[S_4 = \frac{3}{2} \frac{1 - \frac{81}{16}}{1 - \frac{3}{2}}]

[S_4 = \frac{3}{2} \frac{\frac{16}{16} - \frac{81}{16}}{\frac{2}{2} - \frac{3}{2}}]

[S_4 = \frac{3}{2} \frac{-\frac{65}{16}}{-\frac{1}{2}}]

[S_4 = \frac{3}{2} \times \frac{65}{16} \times 2]

[S_4 = \frac{195}{16}]

So, the fourth partial sum of the series is (\frac{195}{16}).

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Answer 2

#sum_1^(n=4)(3/2)^n = 12.1875#

This is geometric progression series of which

first term is # a=3/2=1.5# , common ratio is #r=1.5#
#4 # th partial sum ; i.e #n=4#
Sum # S= a * (r^n-1)/(r-1)= 1.5 * ((1.5^4-1)/(1.5-1))=12.1875#
#sum_1^(n=4)(3/2)^n = 12.1875# [Ans]
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Answer 3

#195/16#

When dealing with a sum, you have a sequence that generates the terms. In this case, you have the sequence

#a_n = (3/2)^n#
Which means that #n#-th term is generates by raising #3/2# to the #n#-th power.
Moreover, the #n#-th partial sum means to sum the first #n# terms from the sequence.
So, in your case, you're looking for #a_1+a_2+a_3+a_4#, which means
#3/2 + (3/2)^2 + (3/2)^3 + (3/2)^4#

You may compute each term, but there is a useful formula:

#sum_{i=1}^n k^i= \frac{k^{n+1}-1}{k-1}#

So, in your case

#sum_{i=0}^4 (3/2)^i= \frac{(3/2)^{5}-1}{3/2-1} = 211/16#
Except you are not including #a_0 = (3/2)^0 = 1# in your sum, so we must subtract it:
#sum_{i=0}^4 (3/2)^i = sum_{i=1}^4 (3/2)^i - 1 = 211/16 - 1 = 195/16#
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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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