**Base Case:** . WTS:
.

By definition, , and since for all such that , it follows that , then , which is what we wanted to show.

**Induction Step:** Assume
for all such that .

WTS: .

By definition, . By hypothesis, . By definition of the -Fibonacci sequence, .

**Base Case:** . WTS:
.

Since , then , which is what we wanted to show.

**Induction Step:** Assume
.

WTS: .

Note that . Now, since for all such that , it follows that , we can apply Theorem 19 to say that . Now, , so we can substitute to get . By hypothesis, , so we can substitute to get , which means that , which is what we wanted to show.

Done by strong mathematical induction on .

**Base Case:** . WTS:
.

By definition of , this is true.

**Induction Step:** Assume
for all
such that
.

WTS: .

Since we know the induction step is true for all such that , we can take the sum of each side as goes from to to say that , which is the same as saying (by definition of the -Fibonacci sequence), which is what we wanted to show.

**Base Case:** . WTS:
.

By Theorem 21, this is true.

**Induction Step:** Assume
.

WTS: .

(by arithmetic). So, by assumption, , and by transitivity, . Now, by Theorem 21, , so by transitivity, , which is what we wanted to show.

Note that from this theorem, the corollary that if , then follows trivially.

**Part 1:** WTS:
.

By the definition of LCM, .

**Part 2:** WTS: For all such that
, it follows that
.

Let , so WTS that for all such that , it follows that . Now, since , we know that and for some . Since and and are positive, we also know that and , so . So, by Theorem 22, we can say that for all , . Similarly, for all , . Thus, for all , , and for all , . Now, if we let and take on all integral values between and (inclusive), we find that for all such that , , and . Now, since and are between and , and by Part 2 of the definition of , we can conclude that, for all such that , then and . Thus, by Theorem 25, for all such that , then .

**Part 3:** WTS:
.

Since , then it follows that . Similarly, . So, by the corollary to Theorem 22, and . Thus, and , so and . Since (which was given), we can use Fact 8 to say that . So, , and thus , which is what we wanted to show.

**Part 4:** WTS: If there exists an such that for all such that
, it follows that
,
and
, then
.

Since for all such that , it follows that , then we can say that for all such that , it follows that (by Fact 16. Also, since , then (by Fact 16). Thus, by Theorem 24, . Similarly, . So, by the definition of LCM, , and by Fact 3, , which is what we wanted to show.

Let , so (by definition). Since for all between and , inclusive, (by definition of ), we know that . Now, if we find that the first directly after s is in position number , then , since and there are s preceding that term, which satisfies the definition of . So, we must examine position numbers , since all sequences of s end in position numbers (by Theorem 23). Now, by Theorem 20, . So, we are looking for the first such that . By definition, this is . So, , or .