Mammoth Memory

Triangles and the `n^(th)` term

`1`   `3`   `6`   `10`   `15`

You can recognise a triangular sequence by picturing it as follows:

This is the easiest way to recognise the triangular sequence 

We can see a pattern emerge

`1^(st)` term `=1` `=1`
`2^(nd)` term `=1+2` `=3`
`3^(rd)` term `=1+2+3` `=6`
`4^(th)` term `=1+2+3+4` `=10`
`5^(th)` term `=1+2+3+4+5` `=15`

This also gives us a clue as to how we find the `n^(th)`  term of any triangular sequence (But note the following only applies when the sequence starts at the number 1).

 

At first, just take one large even set of `n^(th)`  terms

Try `12^(th)` term

The `12^(th)` term would be

`1+2+3+4+5+6+7+8+9+10+11+12=78`

But a mathematician found that if you started by adding the `1^(st)` and last term together and then continue this arrangement then a pattern emerges.

i.e.

By adding the first and last together and continuing on from that a rainbow pattern emerges 

`12^(th)` term`=` `1+12` `=13`  
  `2+11` `=13`  
  `3+10` `=13`  
  `4+9` `=13`  
  `5+3` `=13`  
  `6+7` `=13`  
   
`6times13`
`=78`

So the pattern is `13` and the multiple is `6`.

This works for any even numbered triangular number.

Try `16^(th)` term

`16^(th)`  term`=` `1+16` `=17`  
  `2+15` `=17`  
  `3+14` `=17`  
  `4+13` `=17`  
  `5+12` `=17`  
  `6+11` `=17`  
  `7+10` `=17`  
  `8+9` `=17`  
   
`8times17`
`=136`

So the pattern is `17` and the multiple is `8`.

Now we can form a formula for this

 

Stage 1 find the pattern number

`12^(th)` term `=(12+1)=13`

`16^(th)` term `=(16+1)=17`

So the pattern number `=(n^(th)\ \n\u\m\b\e\r+1)`

 

Stage 2

And the amount of times we repeat this pattern number (the multiple)

`12^(th)` term `=(12)/2=6`

`16^(th)` term `=(16)/2=8`

So the multiple `=(n^(th)\ term)/2`

 

Put stage 1 and stage 2 together we get:

`(n^(th)\ term+1)times(n^(th)\ term)/2`

So the `n^(th)\ term=(n+1)timesn/2`

 

And amazingly this also works for odd triangular numbers too.

 

Example 1

What is the `9^(th)` term of the triangular sequence

Either draw it

Find the 9th term using the rainbow pattern, it works for odd numbers too 

`9^(th)` term`=` `1+9` `=10`  
  `2+8` `=10`  
  `3+7` `=10`  
  `4+6` `=10`  
   
 
    `4times10` `+5=45`

NOTE:

We add `5` because that is the number missed in the series of semi-circles above.

 

or using the formula

`(n+1)timesn/2`

`9^(th)`  term `=(9+1)times9/2`
  `=(10times9)/2`
  `=5times9`
  `=45`

 

 

Example 2

What is the `11^(th)` term of the triangular sequence

Either draw it

Find the 11th term using the rainbow pattern

`11^(th)` term`=` `1+11` `=12`  
  `2+10` `=12`  
  `3+9` `=12`  
  `4+8` `=12`  
  `5+7` `=12`  
   
 
    `5times12` `+6=66`

 NOTE:

We add `6` because that is the number missed in the series of semi-circles above.

 

or using the formula

`(n+1)timesn/2`

`11^(th)`  term `=(11+1)times11/2`
  `=(12times11)/2`
  `=6times11`
  `=66`

 

 

Example 3

What is the `27^(th)` term of the triangular sequence

We are not going to draw this out but we can just rely on the formula.

`(n+1)timesn/2`

`27^(th)`  term `=(27+1)times27/2`
  `=(28times27)/2`
  `=14times27`
  `=378`

 

 

Example 4

What is the `28^(th)` term of the triangular sequence

Again we are not going to draw this out but the formula gives:

`(n+1)timesn/2`

`28^(th)`  term `=(28+1)times28/2`
  `=(29times28)/2`
  `=29times14`
  `=406`

 

(As a point of interest comparing example 3 and the answer 378 and here 406 the difference is 28 which would be correct.)

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