sum of adjacent number in a triangle Ask

Question

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.

Answer 1

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.

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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 $$S_n = 2^{n-1}+2^n-2$$

Answer 2

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!)