George tells you that when variables are in the denominator, the equation becomes unsolvable. "There is a value for x that makes
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S1 = 1
S2 = 0.5
S3 = 0.6667
S4 = 0.625
S5 = 0.6333
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Explanation:
Let
The summation given to us represents the following
There are infinitely many terms to be added.
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The partial sums only care about adding a finite amount of terms.
The partial sum is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say
I'm skipping the steps to compute f(1) since you already have done so.
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The second partial sum is when things get a bit more interesting.
We add the first two terms.
The scratch work for computing f(2) is shown in the diagram below.
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We do the same type of steps for the third partial sum.
The scratch work for computing f(3) is shown in the diagram below.
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Now add the first four terms to get the fourth partial sum.
As before, the scratch work for f(4) is shown below.
I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.
This means something like and
In general, so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.
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Let's use that recursive trick to find
The scratch work for f(5) is shown below.
