# What is the formula for these sequence? ______________1 _____________2 3 ___________4 5 6 _____________... 46 47 48 49 50 51 52 53 54 55

*print*Print*list*Cite

### 3 Answers

You can apply Gauss system, this is a triangle , let's write firsts rows:

1

2 3

4 5 6

7 8 9 10

11 12 13 14 15

Now we concern on the last number of any row giving them a "name" as the first one.

Note than every last number of the n-th row is the sum of the number row with previous ones :

I.e: 3, last number of second row , is sum of 1, (first row) and 2(second row ).So the last number on third row, 6, si sum of 1,2,3, the numbers of the first three rows and so on,

Designed with r, the number of row, you can get the last number of ,with the sum of r numbers row.

Gauss found,when was still a young one, the formula to find sum of r integers in natural sequences.

`S_r= (r^2+r)/2`

Since numbers of every rows running on integer sequence too, they start form the next of previous sequence number sum, to the sum of current row number.

I.e the r row end with `(r^2+r)/2`

Row r+1, start with the next ntegers in natural sequence, that is:

`(r^2+r)/2+1` to the number stands for sum of r rows number:

`((r+1)^2+(r+1))/2` `=((r+1)(r+2))/2`

So the sqeuence you are searching for, is composed by row actually contents all integers in the interval :

`r- row:` `[((r+1)^2-(r-1))/2; ((r+1)(r+2))/2]`

What is the formula for these sequence?

1

2 3

4 5 6

...

46 47 48 49 50 51 52 53 54 55

A sequesnce is defined as

`{S_n}` ,

where`S_n` is the `n^(th)` term of the sequence and have n terms.

First term of the `S_n` is

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

where `a_(n-1) ` is first term of `S_(n-1)` , and `n^(th)` terms of `S_n` is

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

. Also the first term of `S_1` is `a_1=1`

`ie.`

`{S_1}={a_1=1}`

`{S_2}={a_1,a_2}`

`{S_3}={a_1,a_2,a_3}`

`.`

`.`

`.`

`{S_n}={a_1,a_2,.................,a_n}`