sum of adjacent number in a triangle Ask 
Question
Given an Arithmetic Progression of length $n$, construct a right angle triangle with height $n$ and width $n$ with given AP. Now fill the triangle with numbers (formed by sum of adjacent number)

Example
AP = $1,2,3,4,5$
Triangle:
                5
           11   4
       14   7   3
   11   7   4   2
5   4   3   2   1

Problem
Now I have to find sum of all the numbers in that triangle.
In this I am trying to find a formula to get the sum of triangle formed by an AP.
But I am unable to do that.
Someone can help me in this.
 A: Let $S_n$ be the sum of entries in diagonal $n$.
Skipping the edges, notice that each entry in diagonal $n$ appears two times in diagonal $n+1$. And the edge entries increment by the AP common difference, say $d$. Then the sum along each diagonal is:
$$S_{n+1} = 2S_n + 2d$$
Initial condition being $S_1=a$, first term.
You may solve it using any of your favorite methods and get the solution:
$$S_n = a2^{n-1}+d2^n-2d$$
That's the sum of entries in diagonal $n$.
See if you can take this from here.

For the given triangle $d=1$ and $S_1=a=1$, so the recurrence relation is $S_{n+1}=2S_n + 2$, and the slution is $$S_n = 2^{n-1}+2^n-2$$

A: You are looking at a Pascalish triangle of the form
$$\begin{array}{c}
&&&a&&&\\
&&a+d&&a+d&&\\
&a+2d&&2a+2d&&a+2d&\\
a+3d&&3a+4d&&3a+4d&&a+3d\\
&&&\vdots&&&
\end{array}$$
The sequence of row sums is
$$a, 2a+2d,4a+6d,8a+14d,16a+30d,\ldots$$
It's easy to guess that the coefficient of the $a$ is $2^n$ (with $n=0,1,2,\ldots$), and not too much harder to guess that the coefficient of $d$ is $2^{n+1}-2$.  Can you prove that those guesses are correct?
(Minor aside: I had a heck of a time getting TeX to produce that triangle, and it still doesn't look very good -- the entries are a little widely spaced and they aren't properly centered above one another.  If anyone knows the "right" way to produce a Pascal's triangle, I would greatly appreciate it!)
