Hello, outkast32!
There is no formula for #2.
I doubt that your textbook will help you.
It requires some Original Thinking.
2. Let $\displaystyle A \,=\,\{5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$
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:
. . $\displaystyle (5,18),\:(6,17),\:(7,16),\:(8,15),\:(9,14),\:(10,1 3),\:(11,12)$
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:
. . $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 \:=\:\frac{n^2(n+1)^2}{4}$
Verify $\displaystyle S(1)\!:\;1^3 \:=\:\frac{1^2\!\cdot2^2}{4}$ . . . True!
Assume $\displaystyle S(k)\!:\;1^3+2^3+3^3+\cdots+k^3 \;=\;\frac{k^2(k+1)^2}{4}$ .[1]
We must prove $\displaystyle S(k+1)\!:\;1^3+2^3+3^3+\cdots + (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4}$ .[2]
Add $\displaystyle (k+1)^3$ to both sides of [1].
.$\displaystyle 1^2+2^3+3^3+\cdots+k^2+(k+1)^3 \;=\;\frac{k^2(k+1)^2}{4} + (k+1)^3$
Simplify the right side:
. . $\displaystyle \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]$
. . $\displaystyle =\;(k+1)^2\,\frac{(k+2)^2}{4} \;=\;\frac{(k+1)^2(k+2)^2}{4}$
We have proved [2]: .$\displaystyle 1^3+2^3+3^3 + \cdots (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4}$
The inductive proof is complete.