a + ar + ar² + … + arn + …?S of odd positive integers less than 7 in set-builder notation is:|P(A)| = 32 and |P(B)| = 8, then |A × B| is:1·2 + 2·3 + … + n(n+1) = n(n+1)(n+2)/3, which term is added to both sides to go from P(k) to P(k+1)?Σi=02 Σj=03 j equals:(A + B̅)(A + C̅)(B + C).(A + B̅)(A + C̅) = A + B̅C̅, then (A + B̅C̅)(B + C) = AB + AC = A(B + C), which is not among the four printed options. Because no option is correct, this branch was awarded to every student. (Without the complements, (A + B)(A + C)(B + C) = AB + AC + BC, which is option (b).)k is Ak = 2k.n using a single summation.Level k contributes Ak = 2k ancestors. Summing over the levels k = 1 to n:
$$\text{Total} = \sum_{k=1}^{n} 2^{k}$$
Epp, Section 5.1 (Sequences and summation notation).
2n+1 − 2.The terms 2 + 4 + 8 + … + 2n form a geometric series with first term a = 2, common ratio r = 2, and n terms. The geometric-sum formula gives:
$$\sum_{k=1}^{n} 2^{k} = a\cdot\frac{r^{n}-1}{r-1} = 2\cdot\frac{2^{n}-1}{2-1} = 2(2^{n}-1) = 2^{n+1}-2.$$
The same result follows by expansion: 2 + 4 + … + 2n doubles and telescopes to 2n+1 − 2. ∎
Epp, Section 5.2 (Sums of geometric sequences).
n ≥ 1, (2n)! / (2(n−1))! = 4n² − 2n, using the recursive meaning of factorial and no factorial notation in the final answer.First simplify the denominator: 2(n − 1) = 2n − 2, so we have (2n)! / (2n−2)!.
Unfold the top two factors of (2n)! down to (2n−2)!:
$$\frac{(2n)!}{(2n-2)!} = \frac{(2n)(2n-1)\,(2n-2)!}{(2n-2)!} = (2n)(2n-1).$$
The (2n−2)! cancels, leaving (2n)(2n − 1) = 4n² − 2n. ∎
Epp, Section 5.1 (Factorial and the recursive definition).
Z. A = {x ∈ Z | 1 ≤ x ≤ 6}, B = {x ∈ A | x is even}, C = {x ∈ Z | x = 2k+1 for some integer k} (the odd integers).(1) Roster form. A = {1,2,3,4,5,6}, so the even members give B = {2, 4, 6}. The odd integers intersected with A give A ∩ C = {1, 3, 5}.
(2) Set difference and its relationship. A − B = {1, 3, 5} (the members of A not in B). By the definition of set difference,
$$A - B = A \cap B^{c},$$
the elements that are in A and not in B. Here that set also happens to equal A ∩ C = {1,3,5}, since within A "not even" and "odd" describe the same elements.
(3) Is 𝒜 = {B, A−B} a partition of A? Yes. With B = {2,4,6} and A−B = {1,3,5}, all three partition conditions hold:
{2,4,6} and {1,3,5} are non-empty.B ∩ (A−B) = ∅.{2,4,6} ∪ {1,3,5} = {1,2,3,4,5,6} = A.All three hold, so 𝒜 is a valid partition of A. ∎
Epp, Section 6.1 (Set operations, difference, and partitions).
n ≥ 1, 1³ + 2³ + … + n³ = [n(n+1)/2]². Label the basis, hypothesis, and inductive step.Basis (n = 1). LHS = 1³ = 1; RHS = [1·2/2]² = 1. They agree. ✔
Inductive hypothesis. Assume for some k ≥ 1 that Σi=1k i³ = [k(k+1)/2]².
Inductive step. Add (k+1)³ to both sides:
$$\sum_{i=1}^{k+1} i^{3} = \left[\frac{k(k+1)}{2}\right]^{2} + (k+1)^{3} = (k+1)^{2}\left[\frac{k^{2}}{4} + (k+1)\right] = (k+1)^{2}\cdot\frac{k^{2}+4k+4}{4} = \left[\frac{(k+1)(k+2)}{2}\right]^{2}.$$
This is the formula at n = k+1. By the principle of mathematical induction it holds for all n ≥ 1. ∎
Epp, Section 5.3 (Mathematical induction).
(A ∪ B) ∩ (A ∪ Bc) = A algebraically, naming the law used at each step.(A ∪ B) ∩ (A ∪ Bc) = A ∪ (B ∩ Bc) | Distributive law |
= A ∪ ∅ | Complement law |
= A | Identity law |
Hence (A ∪ B) ∩ (A ∪ Bc) = A. ∎
Epp, Section 6.2 (Set identities and Boolean algebra).