Rewrite multiplication of quotient.

Oct 2013
25
0
I'm not sure how I should rewrite this.

n(n-1) * (1/n - 1/n+1)

So I think n(n-1) = n^2 -n
and (1/n - 1/n+1) = 1/n+1
But I'm not sure.

I've got n-1/n+1 as my final answer, but are my intermediate steps right? How should I proceed?
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
Your rewriting of the second factor is incorrect, you want:

\(\displaystyle \frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n(n+1)}=\frac{1}{n(n+1)}\)
 
Oct 2013
25
0
Ok. Thank you!
So could you tell me how should I get: n+1-n/n(n+1)
Which steps should I follow?
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
As with numeric fractions, we need to get a common denominator before we can add or subtract. The two denominators have no factors in common, and so the lowest common denominator is the product of the two. Thus, the first term needs to be multiplied by \(\displaystyle 1=\frac{n+1}{n+1}\) and the second term by \(\displaystyle 1=\frac{n}{n}\). This will give both terms the common denominator of \(\displaystyle n(n+1)\):

\(\displaystyle \frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}\cdot\frac{n+1}{n+1}-\frac{1}{n+1}\cdot\frac{n}{n}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\)

Now we have the same denominator, and we may apply the identity:

\(\displaystyle \frac{a}{c}\,\pm\,\frac{b}{c}=\frac{a\pm b}{c}\)

to get:

\(\displaystyle \frac{n+1-n}{n(n+1)}\)

Now, combine like terms in the numerator:

\(\displaystyle \frac{1}{n(n+1)}\)