Find the solution to the recurrence relation and prove the correctness of your solution using mathematical induction. a_n=a_{n-1}+3, n >= 1; a_0=1

Let A, B, and C be sets. Show that ${\displaystyle (A-B)-C=(A-C)-(B-C)}$

How many elements are in the set { 0, { { 0 } }?

Suppose that A is the set of sophomores at your school and B is the set of students in discrete mathematics at your school. Express each of these sets in terms of A and B.
a) the set of sophomores taking discrete mathematics in your school
b) the set of sophomores at your school who are not taking discrete mathematics
c) the set of students at your school who either are sophomores or are taking discrete mathematics
Use these symbols: ${\displaystyle \cap \cup }$

Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.

Direct Proof Discrete Math.
Show that if n is an odd integer, then $n^2$ is odd.
Proof : Assume that n is an odd integer. This implies that there is some integer k such that $n=2k+1$. Then $n^2=(2k+1{)}^2=4k^2+4k+1=2(2k^2+2k)+1$. Thus, $n^2$ is odd.
Why does the solution assume n to be $2k+1$?
How do you know $n^2$ is odd based on $2(2k^2+2k)+1$?
I don't see how $2k+1$ for n and 2(2k2+2k)+1 for $n^2$ means an odd integer. Some clarification would be helpful.

Discrete Math Proofs
Are these steps for finding the solutions of $\sqrt{x+3}=3-x$correct?
1. $\sqrt{x+3}=3-x$ is given;
2. $x+3=x^2-6x+9$, obtained by squaring both sides of (1);
3. $0=x^2-7x+6$, obtained by subtracting $x+3$ from both sides of (2);
4. $0=(x-1)(x-6)$, obtained by factoring the right-hand side of (3);
5. $x=1$ or $x=6$, which follows from (4) because $ab=0$ implies that $a=0$ or $b=0$.

In how many ways can we distribute 2 types of gifts?
The problem: In how many ways can we distribute 2 types of gifts, m of the first kind and n of the second to k kids, if there can be kids with no gifts?
From the stars and bars method i know that you can distribute m objects to k boxes in $(\genfrac{}{}{0ex}{}{m+k-1}{k-1})$ ways. So in my case i can distribute m gifts to k kids in $(\genfrac{}{}{0ex}{}{m+k-1}{k-1})$ ways, same for n gifts i can distribute them in $(\genfrac{}{}{0ex}{}{n+k-1}{k-1})$ ways. So now if we have to distribute m and n gifts we can first distribute m gifts in $(\genfrac{}{}{0ex}{}{m+k-1}{k-1})$ ways, then n gifts in $(\genfrac{}{}{0ex}{}{n+k-1}{k-1})$ ways, so in total we have:
$(\genfrac{}{}{0ex}{}{m+k-1}{k-1})\cdot (\genfrac{}{}{0ex}{}{n+k-1}{k-1})\text{ways.}$.
Is my reasoning correct?
What about when we have to give at least 1 gift to each kid, can we do that in
$(\genfrac{}{}{0ex}{}{m-1}{k-1})\cdot (\genfrac{}{}{0ex}{}{n+k-1}{k-1})+(\genfrac{}{}{0ex}{}{n-1}{k-1})\cdot (\genfrac{}{}{0ex}{}{m+k-1}{k-1})\text{ways?}$