How do I get the equation for the 10th, 20th and nth members in the sequence?

The sequence is composed of triangles to make a pyramid;

1. 1 trangle; 3 matchsticks

2. 4 triangles; 9 matchsticks

3. 9 triangles; 18 matchsticks

Looks like this: http://nrich.maths.org/88

How do I find the pattern to make the equation to find the 10th, 20th, and nth members of the sequence?

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In order for us to know how to obtain terms that are far down these lists of numbers, we need to develop a formula that can be used to calculate these terms. If we were to try and find the 20th term, or worse to 2000th term, it would take a long time if we were to simply add a number -- one at a time -- to find our terms.

If a 5-year-old was asked what the 301st number is in the set of counting numbers, we would have to wait for the answer while the 5-year-old counted it out using unnecessary detail. We already know the number is 301 because the set is extremely simple; so, predicting terms is easy. Upon examining arithmetic sequences in greater detail, we will find a formula for each sequence to find terms.

- Let's examine sequence A so that we can find a formula to express its nth term.
If we match each term with it's corresponding term number, we get:

**n**1 2 3 4 5 . . .**Term**5 8 11 14 17 . . .The fixed number, called the common difference (d), is 3; so, the formula will be an = dn + c or an= 3n + c, where c is some number that must be found.

For sequence A above, the rule an = 3n + c would give the values...

3×1 + c = 3 + c

3×2 + c = 6 + c

3×3 + c = 9 + c

3×4 + c = 12 + c

3×5 + c = 15 + cIf we compare these values with the ones in the actual sequence, it should be clear that the value of c is 2. Therefore the formula for the nth term is...

an = 3n + 2.

Now if we were asked to find the 37th term in this sequence, we would calculate for a37 or 3(37) + 2 which is equal to 111 + 2 = 113. So, a37 = 113, or the 37th term is 113. Likewise, the 435th term would be a435= 3(435) + 2 = 1307.

- Let's take a look at sequence B.
**n**1 2 3 4 5 . . .**Term**26 31 36 41 46 . . .The fixed number, d, is 5. So the formula will be an = dn + c or an= 5n + c .

For the sequence above, the rule an= 5n + c would give the values...

5×1 + c = 5 + c

5×2 + c = 10 + c

5×3 + c = 15 + c

5×4 + c = 20 + c

5×5 + c = 25 + cIf we compare these values with the numbers in the actual sequence, it should be clear that the value of c is 21. Therefore, the formula for the nth term is...

an = 5n + 21.

If we wanted to calculate the 14th term, we would calculate for

a14 = 5(14) + 21 = 70 + 21 = 91. If we needed the 40th term, we would calculate a40= 5(40) + 21 = 200 + 21 = 221. The general formula is very handy.

- Now let's do the third and final example....
**n**1 2 3 4 5 . . .**Term**20 18 16 14 12 . . .The common difference is -2. So the formula will be -2n + c, where c is a number that must be found.

For sequence C, the rule -2n + c would give the values...

-2×1 + c = -2 + c

-2×2 + c = -4 + c

-2×3 + c = -6 + c

-2×4 + c = -8 + c

-2×5 + c = -10 + cIf we compare these values with the numbers in the actual sequence, it should be clear that the value of c is 22. Therefore, the formula for the nth term is...

an = -2n + 22.

If for some reason we needed the 42nd term, we would calculate for a42 = -2(42) + 22 = -84 + 22 = -62. Similarly, a90 = -2(90) + 22 = -180 + 22 = -158.

**Sources:**

Look how the pyramid is constructed.

The pyramid is constructed inside an equilateral triangle of size n sticks. It is filled with small equilateral triangles of size 1.

If `a_n ` is the number of small triangles in the big triangle of side n.

`b_n` is the number of sticks needed to fill the big triangle of side n.

It build the next step, we just need to add n+1 small upward traingles of side 1

/_\ /_\ ... /_\

Which means we add 3(n+1) sticks.

Therefore `b_(n+1)=b_n+3(n+1)`

How many small triangles were added?

(n+1) upward and n downward. i.e 2n+1 in total.

`a_(n+1)=2n+1+a_n=2(n+1)-1+a_n`

Let's try to find a expression of `b_n` as a function of n

`b_n=3+3*2+3*3+3*4+....+3*n=3(1+2+...+n)=3*n(n+1)/2`

b_n=3n(n+1)/2

Let's find `a_n`

`a_n=2*1-1+2*2-1+2*3-1+...2*n-1=2(1+2+3+...+n)-1-1-...-1`

`a_n=2n(n+1)/2-n=n(n+1)-n=n^2.`

**Therefore, at the level n there are** `3n(n+1)/2` **sticks and** `n^2 ` **triangles.**

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