# 3+6+9+ …………………+3n=(3n(n+1))/2 I assume you are asked to show that the finite sum (this is a series; the sum of the elements of some sequence):

3+6+9+...+3n=(3n(n+1))/2

We can rewrite the equation as :

`3(1+2+3+...+n)=3((n(n+1))/2)`

Which is equivalent to 1+2+3+...+n=n(n+1)/2.

The nth partial sums are called the triangular numbers. 1,1+2=3,1+2+3=6,1+2+3+4=10, etc...

Each of 1, 3, 6, 10, 15, 21,... can be drawn as a set of dots in an equilateral triangle (1 being a degenerate case.)

(1) For each partial sum we can draw an nx(n+1) rectangle. (Note that one of n or (n+1) will be even, so the product n(n+1) is even.) Drawing a diagonal line so that there are the same number of dots on each side yields "L" shaped figures with 1, 3, 6, 10, etc... dots.

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Etc. These triangles have 1, 1+2, 1+2+3, etc. dots in them and can be seen to be 1/2 the number of dots in a nx(n+1) rectangle. So 1+2+3+...+n =(n(n+1))/2.

So 1+2+3+...+n=(n(n+1))/2 and 3+6+9+...+3n=(3n(n+1))/2 as required.

(2) A possible apocryphal story about the mathematician Karl F. Gauss has the young Gauss in a class when the teacher asks the students to add up the numbers from 1 to 100 (perhaps hoping for some quiet time.) Young Gauss almost immediately has the answer. His method:

Write the sum in one direction. Beneath write the sum in reverse order and add the two series.

1+2+3+...+98+99+100
100+99+98+...+3+2+1
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101+101+101+...+101+101+101

There are 100 101's, which is 10100. But that is adding the series twice, so divide by 2 to get 5050, which is the answer.

Thus, for a finite arithmetic series (each term in the underlying sequence differs from the previous term by the same common difference), the sum can be found by:

`S_(n)=(n(a_1+a_n))/2` where n is the number of terms, a(1) is the first term and a(n) is the nth term. For Gauss, we have `S_(100)=(100(1+100))/2`

For the arithmetic series 3+6+9+...+3n, we get:

`S_(n)=(n(3+3n))/2=(3n(1+n))/2` as required.

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