1. The natural numbers are written in a triangle as shown below
Prove that the sum of the numbers in the nth row is n(n^2)/2
---For this one I figured out that if you subtract the sum of all the numbers before the nth row from the sum of all the numbers including the nth row the you'll get the sum of the nth row, I just cant figure out how to put that in math terms to prove the statement.

2. In the diagram below, D is the midpoint of AB, and T trisects AC. Segment DT is extended to meet BC extended at U.
a) Prove that CU is the same length as BC
b) Prove that TU is twice as long as DT
c) Let E be the midpoint of AT. Segment ED is extended to meet BC extended at V. Prove that BV is the same length as BC.
----I've been looking at this one for a while now, I've proved that DU is a median so the area of ADU is equal to the area of BDU and I've been trying to manipulate values and triangle sections but I just cant seem to get it to work

3.To construct a regular octagon, construct a square, and locate its centre, O. The construct two arcs through O with centres at opposite vertices. Using the other vertices, draw similar arcs through O. Join the points located on the square to form an octagon. Prove the octagon is a regular octagon (all the sides have the same length and angles are equal).

I've pretty much finished this one, I've proved that the all the sides that are the hypotenuse of the corner triangles are equal and that all the sides that are on the edge of the square are equal. What I'm stuck on is how to prove the diagonal sides are equal to the horizontal/vertical sides. I've tried using a coordinate system and distances, but it didnt seem to work out. I've proved the angles are equal already.

2. Hello, omgmath!

The natural numbers are written in a triangle as shown below.

Prove that the sum of the numbers in the nth row is: $\displaystyle \frac{n(n^2+1)}{2}$

$\displaystyle \begin{array}{c} 1 \\ 2\quad3 \\ 4\quad5\quad6 \\ 7\quad8\quad9\quad10 \\ 11\;\;12\;\;13\;\;14\;\;15\end{array}$

Note that each rows ends with a Triangular Number: 1, 3, 6, 10, 15, 21, . . .

The $\displaystyle k^{th}$ triangular number is: .$\displaystyle T_k\;=\;1 + 2 + 3 + \cdots + k\;=\;\frac{k(k+1)}{2}$

We want the sum of the integers up to $\displaystyle T_n$
. . minus the sum of integers up to $\displaystyle T_{n-1}$

So we have: .$\displaystyle \frac{(T_n)(T_n + 1)}{2} \,- \,\frac{(T_{n-1})(T_{n-1} + 1)}{2} \;=\; \frac{\left[\frac{n(n+1)}{2}\right]\left[\frac{n(n+1}{2} + 1\right]}{2} \,- \,\frac{\left[\frac{n(n-1)}{2}\right]\left[\frac{n(n-1)}{2} + 1\right]}{2}$

. . which simplifies to: .$\displaystyle \frac{n(n^2+1)}{2}$

3. Hello, omgmath!

3.To construct a regular octagon, construct a square, and locate its centre, O.
Then construct two arcs through O with centres at opposite vertices.
Using the other vertices, draw similar arcs through O.
Join the points located on the square to form an octagon.
Prove the octagon is a regular octagon.
We have a square $\displaystyle ABCD$ with center $\displaystyle O$ and side $\displaystyle 2a.$
Code:
      A     Q       P     B
* - - o - - - o - - *
| *               * |
|   *           *   |
R o     *       *     |
|       *   *       |
|         *         | 2a
|       * O *       |
|     *       *     |
|   *           *   |
| *               * |
* - - - - - - - - - *
D        2a         C

Using center $\displaystyle A$ and radius $\displaystyle r = AO$, locate point $\displaystyle P$.
Using center $\displaystyle B$ and radius $\displaystyle r = BO$, locate point $\displaystyle Q$.
Using center $\displaystyle D$ and radius $\displaystyle r = DO$, locate point $\displaystyle R$.

The radii of all the arcs are equal to: .$\displaystyle r \:=\:AO \:=\:a\sqrt{2}$
. . Hence: .$\displaystyle AP \:=\:BQ \:=\:DR \:=\:a\sqrt{2}$
. . And: .$\displaystyle AQ \:= \:BP \:= \:AR \:= \:(2-\sqrt{2})a$

Then: . $\displaystyle PQ \:=\:2a(\sqrt{2}-1)$
. . This is the length of the four vertical and horizontal sides of the octagon.

We have isosceles right triangle $\displaystyle QAR$ with legs $\displaystyle (2-\sqrt{2})a$

Its hypotenuse is given by: .$\displaystyle QR^2\;=\;[(2-\sqrt{2})a]^2 + [(2-\sqrt{2})a]^2$

. . Then: .$\displaystyle QR^2\;= \;4a^2(3 - 2\sqrt{2}) \;=\;4a^2(\sqrt{2}-1)^2$

So: .$\displaystyle PQ \;=\;2a(\sqrt{2}-1)$
. . This is the length of the four diagonal sides of the octagon.

Hence, the octagon has eight equal sides.

Since each of the four corners has the same isosceles right triangle,
. . all the vertices have 135° angles.
Hence, the octagon has eight equal angles.

Therefore, we have a regular octagon.