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We know that an is for n = 0, 1, 2, 3, ...

We shall now find out what an is when n is negative.

We know that    and we used his to find a0 (we also implicitly used this to find a1).

We shall now use this to find an when n is negative.

     

 

     

 

     

 

Similarly,

     

and

     

etc.

     

We could also have derived this in other ways.

One way is by saying that

     

and

     a-n = (a-1)n

Another way is by noting that

     an x a-n = an - n

                  = a0

                  = 1

Thus a-n x a-n = 1

     So  


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Question 1/5

1. (3a)-4 is

34a4







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