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I will be forever in debted if you can help me with these. My first discrete math class.

you have to show some work. These are all straightforward problems you should be able to do after reading your textbook.

Hello, outkast32!

There is no formula for #2.
It requires some Original Thinking.

2. Let $A \,=\,\{5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$
How many numbers must be chosen to ensure that two of them add up to 23?

Note that there are 7 Pairs that add up to 23:
. . $(5,18),\:(6,17),\:(7,16),\:(8,15),\:(9,14),\:(10,1 3),\:(11,12)$

To have a sum of 23, your choices must include one of these Pairs.

Now consider the worst-case scenario: none of the Pairs.
If you select 7 numbers, one from each Pair,
. . you will not have a sum of 23.

When you choose an 8th number, you will have a Pair.

Therefore, you must draw 8 numbers.

Hello, outkast32!

The first problem (unnumbered) is not true.

It should be stated like this:

Using mathematical induction, prove that:

. . $1^3 + 2^3 + 3^3 + \cdots + n^3 \:=\:\frac{n^2(n+1)^2}{4}$

Verify $S(1)\!:\;1^3 \:=\:\frac{1^2\!\cdot2^2}{4}$ . . . True!

Assume $S(k)\!:\;1^3+2^3+3^3+\cdots+k^3 \;=\;\frac{k^2(k+1)^2}{4}$ .[1]

We must prove $S(k+1)\!:\;1^3+2^3+3^3+\cdots + (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4}$ .[2]

Add $(k+1)^3$ to both sides of [1].

. $1^2+2^3+3^3+\cdots+k^2+(k+1)^3 \;=\;\frac{k^2(k+1)^2}{4} + (k+1)^3$

Simplify the right side:

. . $\frac{k^2(k+1)^2}{4} + (k+1)^3 \;=\(k+1)^2\left[\frac{k^2}{4} + (k+1)\right] \;=\;(k+1)^2\left[\frac{k^2+4k+4}{4}\right]$

. . $=\;(k+1)^2\,\frac{(k+2)^2}{4} \;=\;\frac{(k+1)^2(k+2)^2}{4}$

We have proved [2]: . $1^3+2^3+3^3 + \cdots (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4}$

The inductive proof is complete.