6/(n raise to 3 - n)

Question

6/(n³-n)∀nä∈Ζ

Answer 1

Hi Michael,

 

Solving this problem requires the use of exponent rules and how to work with fractions.

 

First, let's take a look at the denominator: n^(3-n)

We can apply our knowledge of exponent division to get: (n^3)/(n^n)

 

So we have 6/[(n^3)/(n^n)].

 

Using the basics of fractions, we can rewrite this as (6*n^n)/(n^3)

 

Apply exponent division again, and we've gotten rid of the fraction to get: 6n^(n-3)

This is the simplified version of the original expression.

 

You can check your work by choosing numbers for n and finding the value of the original and simplified expressions.

 

Do a search for "exponent rules" and "properties of fractions" for some additional information and practice problems. You might consider making a study guide with the exponent rules and some examples.

 

Good luck!

Katherine

Answer 2

Is your question to prove via mathematical induction that n³-n is divisible by 6 for all integers n?

 

If so, we shall first prove the above for the non-negative integers.  First, we begin with a base case; let us take that as n=0.  n=0 clearly satisfies the desired property since 0³-0=0, which is divisible by 6.

 

Secondly, let us appeal to the inductive hypothesis: assume that for an integer n, n³-n divisible by 6.  Then we wish to show the integer immediately after n also satisfies the property: (n+1)³-(n+1) is divisible by 6.

 

The above can be shown by expanding the above expression: (n+1)³-(n+1) = n³+3n²+3n+1-n-1= (n³-n)+(3n²+3n)=(n³-n)+3n(n+1).  n³-n is divisible by 6 by assumption.  Also, it is clear that 3n(n+1) is divisible by 6.

 

The second claim can be seen by the fact that since n is an integer, either n or n+1 will be odd (and correspondingly either n+1 or n will be even).  Thus, representing the even number as 2N for some integer N, n(n+1) can also be written as 2N*(2N+1) (if n is even) or 2N*(2N-1) (if n is odd).  Thus 3n(n+1)=3*2N*(2N+ or - 1)=6N(2N+ or - 1), which is clearly divisible by 6 [6N(2N + or - 1)/6=N(2N+ or - 1)].

 

Since the sum of two numbers are divisible by 6, the sum itself will be divisible by 6, making (n+1)³-(n+1) divisible by 6.

 

Having shown the seed case, and having shown that an integer satisfying the above property implies that the integer after that also satisfies the above property, we have proven the claim for all positive integers.

 

Now, assuming n is positive then -n is negative.  Insert -n into the expression: (-n)³-(-n)=-n³+n=-(n³-n), which is simply the negative of the expression already dealt with before.  Having proven n³-n is divisible by 6, its negative must also be divisible by 6.  This proves the claim for all negative integers.

 

Q.E.D.