Why was the term rearranged like that?

There were NO instructions? What is the title of the section of the text where this appeared.

Why isn't this just addition of fractions - finding a common denominator?

perhaps the author is demonstrating that (n+1) < (n+1)(n+2)?
 
LazeFox said:
View attachment 28536
How do I get this solution? I'm not sure as to what to do here
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If you're really asking "why", as in the title, we'd need to know the context to see the goal.

But for "how", as you ask here, they are just combining the fractions using a common denominator:

[math]1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)} = 1-\left(\frac{1}{n+1}-\frac{1}{(n+1)(n+2)} \right) =\\ 1-\left(\frac{1(n+2)}{(n+1)(n+2)}-\frac{1}{(n+1)(n+2)} \right) = 1-\frac{(n+2)-1}{(n+1)(n+2)}[/math]
 
mikewill54 said:
Why does the + become a - inbetween the two fractions?
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Hi Mike. There are different ways to simplify the given expression algebraically. Dr. Peterson chose to associate the two fractions and then subtract them as a group, but he could have used a different approach, instead. When doing math, there are many personal choices available.

By the way, the result in the op is not fully simplified. However, we can't say whether it ought to be, without any context.

?
 
mikewill54 said:
I meant the following
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Your meaning is clear, Mike. I'd just pointed out that it's not necessary to start by changing that arithmetic operator. There are often many differing ways in beginning algebra to arrive at a particular result.

mikewill54 said:
I can see what you’ve don’t now.
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Your meaning in that statement is not clear, Mike.

"I can see what you have do not now"

? :confused:
 
The equation started with 1 -1/(n+1) + 1/(n+1)(n+2)
So it turns into 1 (-1/(n+1) )+ 1/(n+1)(n+2)
So the answer will be (1 - (n+2))/(n+1)(n+2)
 
The equation started with 1 -1/(n+1) + 1/(n+1)(n+2)
So it turns into 1 +(-1/(n+1) )+ 1/(n+1)(n+2)
So the answer will be 1-(1 - (n+2))/(n+1)(n+2)
There was a mistake I did
 
Sorry again correction must be
The equation started with 1 -1/(n+1) + 1/(n+1)(n+2)
So it turns into 1 +(-1/(n+1) )+ 1/(n+1)(n+2)
So the answer will be 1+(1 - (n+2))/(n+1)(n+2)
 
AbdelRahmanShady said:
Isn’t 1 - a + b = 1 + -a + b
How did you know he meant 1 -(a+b)
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1 + -a + b is not a valid expression, please clarify what you mean.

Here's the explanation:
1 - a + b means we want to subtract a, then add b.
What if we want to take care of a and b first, then subtract the result from 1?
Would it be 1-(a+b)?
No. This means we are subtracting both a and b from 1, do you see it?
b is added to the total. Therefore it should be subtracted from the amount we are subtracting from 1: 1 - (a-b).

Another way:
1-a+b = 1 + (-1)a + b = 1 + (-1)a - (-1)b.
Factoring out (-1): 1 + (-1)(a - b) = 1 - (a - b).
 
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