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The number pattern is

$$1,3,6,10,15,\dots$$

Find formula for this pattern. And thus find the 50th pattern.

I have problems in trying to come out with these formulas, is there a way to see patterns? Let's say if the pattern is changed, is the method still the same? I need advice on how to solve the above pattern as well as what is the method!

Thanks in advance !

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4 Answers

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In these types of "puzzles" there are always several (in fact, infinitely many!) possible answers, but a simple one would be to notice that the difference between each successive term increases with one for each term, i.e.

$$a_{n+1}=a_n+(n+1), \quad \text{with }a_1=1.$$

This can be solved to give $$a_n=\frac{n(n+1)}{2},$$

which is the sum of the natural numbers up until and including $n$ (as Jan also noticed).

These is no method in general, though there are a few things one should try first: Looking at the difference for the first few terms is one of them. Another would be to look for a pattern in all even terms (or odd terms).

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$$\sum_{m=1}^{n}m=\frac{n(n+1)}{2}$$

So, we get:

$$1,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,153,171,190,210,\dots$$

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I suggest you look up 'the method of finite differences' for general tips on solving these sorts of sequences.

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The set $1, 3, 6, 10, 15, ...$ has a obvious pattern:

$ 1+(1+1)=3$

$ 3+(2+1) = 6$

$ 6+(3+1) = 10$

$ 10+(4+1) = 15$

This is the triangular number sequence.

See Triangular number. If you want to find the number of dots, flip a equilateral triangle so that it matches up with the first one. It forms a rectangle with width $ 1 + $triangle length and the height the same as the length of the triangle. The area of the rectangle is $n(n+1)$ where $n$ is the length of the triangle. That means that the number of dots (or the area) of the equilateral triangle is

$$\frac{n(n+1)}{2}$$

The sequence of the number of dots in a triangle is the same as the sequence that you asked for. Easily plug in $n$ for $50$ and you get:

$$ \frac{50(50+1)}{2} = 1275$$

for the answer. It is the same way ($\frac{n(n+1)}{2}$)for any number greater than $1$.

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