# 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...

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?

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.**

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.