Tuesday

October 13, 2015
Posted by **jaznai** on Tuesday, September 12, 2006 at 11:56pm.

1+2+...+80

I'm not sure what the question is, to add up the numbers from 1 to 80? If it is this is an extremely old problem.

If you use the progression

o

1 dot____

..o

..oo

3 dots___

.....o

.....oo

.....ooo

.....oooo

4dots____

the object is to find the total number of dots (or o's here); then the first sum is 1 the second 1+2, the third 1+2+3, etc.

There are a number of ways to show what this sum is. Here is the method attributed to Gauss.

Write the sum forwards and then backwards beneath it like this:

1 + 2 +3+...+79+80=sum

80+79+78+...+ 2 +1=sum

Now add them together to get

81+81+81+...+81+81=2*sum

The LHS is 80*81 so sum=(80*81)/2=40*81

In general to add the numbers 1+2+3+...+n the sum is (n*(n+1))/2. There are other way to prove this too. If you had some other question in mind please post it.

Not quite sure whether you are merely looking for the sum of the numbers from 1 to 80 or the sum of the triangular numbers. The sum of the numbers from 1 to n is given by S = n(n + 1)/2.

The following will familiarize you with the triangular numbers.

Triangular Numbers

The number of dots, circles, spheres, etc., that can be arranged in an equilateral or right triangular pattern is called a triangular number. The 10 bowling pins form a triangular number as do the 15 balls racked up on a pool table. Upon further inspection, it becomes immediately clear that the triangular numbers, T1, T2, T3, T4, etc., are simply the sum of the consecutive integers 1-2-3-4-.....n or Tn = n(n + 1)/2, namely, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66,78, 91, etc.

Triangular numbers are the sum of the balls in the triangle as defined by Tn = n(n + 1)/2.

Order.n...1........2...............3.....................4.............................5....................6.....7.....8.....9

.............O.......O...............O....................O............................O

....................O...O.........O...O...............O...O.......................O...O

..................................O...O...O.........O....O....O...............O....O....O

......................................................O....O...O....O.........O.....O...O....O

.................................................................................O....O....O....O....O

Total......1........3.................6....................10..........................15..................21...28...36...45...etc.

The sum of a series of triangular numbers from 1 through Tn is given by S = (n^3 + 3n^2 + 2n)/6.

After staring at several triangular and square polygonal number arrangements, one can quickly see that the 1st and 2nd triangular numbers actually form the 2nd square number 4. Similarly, the 2nd and 3rd triangulars numbers form the 3rd square number 9, and so on. By inspection, one can see that the nth square number, Sn, is equal to Tn + T(n - 1) = n^2. This can best be visualized from the following:

.........Tn - 1...3...6...10...15...21...28...36...45...55...66...78...91

.........T(n - 1)........1...3....6....10...15...21...28...36...45...55...66...78

.........Sn.........1...4...9....16...25...36...49...64...81..100.121.144.169

A number cannot be triangular if its digital root is 2, 4, 5, 7 or 8.

Some interesting characteristice of Triangular numbers:

The numbers 1 and 36 are both square and triangular. Some other triangular squares are 1225, 41,616, 1,413,721, 48,024,900 and 1,631,432,881. Triangular squares can be derived from the series 0, 1, 6, 35, 204, 1189............Un where Un = 6U(n - 1) - U(n - 2) where each term is six times the previous term, diminished by the one before that. The squares of these numbers are simultaneously square and triangular.

The difference between the squares of two consecutive rank triangular numbers is equal to the cube of the larger numbers rank.

Thus, (Tn)^2 - (T(n - 1))^2 = n^3. For example, T6^2 - T5^2 = 441 - 225 = 216 = 6^3.

The summation of varying sets of consecutive triangular numbers offers some strange results.

T1 + T2 + T3 = 1 + 3 + 6 = 10 = T4.

T5 + T6 + T7 + T8 = 15 + 21 + 28 + 36 = 100 = 45 + 55 = T9 + T10.

The pattern continues with the next 5 Tn's summing to the next 3 Tn's followed by the next6 Tn's summing to the next 4 Tn's, etc.

The sum of the first "n" cubes is equal to the nth triangular number. For instance:

n............1.....2.....3.....4.......5

Tn..........1.....3.....6....10.....15

n^3.........1 + 8 + 27 + 64 + 100 = 225 = 15^2

Every number can be expressed by the sum of three or less triangular numbers, not necessarily different.

1 = 1, 2 = 1 + 1, 3 = 3, 4 = 3 + 1, 5 = 3 + 1 + 1, 6 = 6, 7 = 6 + 1, 8 = 6 + 1 + 1, 9 = 6 + 3, 10 = 10, etc.

Alternate ways of finding triangular squares.

From Tn = n(n + 1)/2 and Sn = m^2, we get m^2 = n(n + 1)/2 or 4n^2 + 4n = 8m^2.

Adding one to both sides, we obtain 4n^2 + 4n + 1 = 8m^2 + 1.

Factoring, we find (2n + 1)^2 = 8m^2 + 1.

If we allow (2n + 1) to equal "x" and "y" to equal 2m, we come upon x^2 - 2y^2 = 1, the famous Pell Equation.

We now know that the positive integer solutions to the Pell equation, x^2 - 2y^2 = +1 lead to triangular squares. But how?

Without getting into the theoretical aspect of the subject, sufficeth to say that the Pell equaion is closely connected with early methods of approximating the square root of a number. The solutions to Pell's equation, i.e., (x,y), often written as (x/y) are approximations of the square root of D in x^2 - 2y^2 = +1. Numerous methods have evolved over the centuries for estimating the square root of a number.

Diophantus' method leads to the minimum solutions to x^2 - Dy^2 = +1, D a non square, by setting x = my + 1 which leads to y = 2m/(D - m^2).

From values of m = 1.......n, many rational solutions evolve.

Eventually, an integer solution will be reached.

For instance, the smallest solution to x^2 - 2y^2 = +1 derives from m = 1 resulting in x = 3 and y = 2 or sqrt(2) ~= 3/2..

Newton's method leads to the minimum solution sqrt(D) = sqrt(a^2 + r) = (a + D/a)/2 ("a" = the nearest square) = (3/2).

Heron/Archimedes/El Hassar/Aryabhatta obtained the minimum solution sqrt(D) = sqrt(a^2 +-r) = a +-r/2a = (x/y) = (3/2).

Other methods exist that produce values of x/y but end up being solutions to x^2 - Dy^2 = +/-C.

Having the minimal solutions of x1 and y1 for x^2 - Dy^2 = +1, others are derivable from the following:

(x + ysqrtD) = (x1 + y1sqrtD)^n, n = 1, 2, 3, etc.

Alternitive approach

Given x = p and y = q satifying x^2 - 2y^2 = +1, we can write (x + sqrtD)(x - sqrtD) = 1.

x = [(p + qsqrt(2))^n + (p - qsqrt(2))^n]/2

y = [(p + qsqrt(2))^n - (p - qsqrt(2))^n]/(2sqrt(2))

Having the minimum solution of x = 3 and y = 2, the next few solutions derive from n = 2 and 3 where x = 17, y = 12, x = 99 and y = 70 respectively.

Alternative approach

Subsequent solutions can also be obtained by means of the following:

x^2 - 2y^2 = +1 can be rewritten as x^2 - 2y^2 = (x + yqrt(2)(x - ysqrt(2)) = +1.

Using the minimum solution of x = 3 and y = 2, we can now write

.................(3 + sqrt(2))^2(3 - sqrt(2))^2 = 1^2 = 1

.................(17 + 12sqrt(2))(17 - 12sqrt(2)) = 1

.................289 - 2(144) = 17^2 - 2(12)^2 = 1 the next smallest solution.

The next smallest solution is derivable from

.................(3 + sqrt(2))^3(3 - sqrt(2))^3 = 1^2 = 1 which works out to

.................(99 + 70sqrt(2))(99 - 70sqrt(2)) = 1 or

.................99^2 - 2(70)^2 = 1.

Similarly, (3 + sqrt(2))^4(3 - sqrt(2))^4 = 1^2 = 1 leads to

.................(577 + 408sqrt(2))(577 - 408sqrt(2)) = 1 and

.................577^2 - 2(408)^2 = 1.

Regardless of the method, we ultimately end up with the starting list of triangular squares.

..x........y........n.......m........Tn = Sm^2

..3........2........1........1..............1

.17......12.......8........6..............36

.99......70......49......35...........1225

577....408....288.....204.........41,616 etc.