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1 answer:
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Prove:
Using mathemetical induction:
P(n) =
for n=1
P(n) = = 6
It is divisible by 2 and 3
Now, for n=k,
P(k) =
Assuming P(k) is divisible by 2 and 3:
Now, for n=k+1:
P(k+1) =
P(k+1) =
P(k+1) =
Since, we assumed that P(k) is divisible by 2 and 3, therefore, P(k+1) is also
divisible by 2 and 3.
Hence, by mathematical induction, P(n) = is divisible by 2 and 3 for all positive integer n.
Answer:
A rational expression that has the nonpermissible values and
is
.
Step-by-step explanation:
A rational expression has a nonpermissible value when for a given value of , the denominator is equal to zero. In addition, we assume that both numerator and denominator are represented by polynomials, such that:
(1)
Then, the factorized form of must be:
(2)
If we know that , then the rational expression is:
(3)
A rational expression that has the nonpermissible values and
is
.