3+6+9+ …………………+3n=(3n(n+1))/2
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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...
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