What is the 100th term of sequence:
1,3,6,10,15,....
Write a explicit formula

- Karon86

What is the 100th term of sequence:
1,3,6,10,15,....
Write a explicit formula

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- schrodinger

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- jim_thompson5910

|dw:1443915861353:dw|

- jim_thompson5910

the jump from 1 to 3 is +2
|dw:1443915905024:dw|

- jim_thompson5910

the jump from 3 to 6 is +3
|dw:1443915932403:dw|

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## More answers

- jim_thompson5910

the jump from 6 to 10 is +4
|dw:1443915954079:dw|

- jim_thompson5910

the jump from 10 to 15 is +5
|dw:1443915976631:dw|

- jim_thompson5910

so what you can do is keep this pattern up until you have generated 100 terms
OR
you can find an explicit closed formula so you can avoid using recursion (recursion is long and tedious for a lot of terms)

- Karon86

I want an explicit formula

- jim_thompson5910

I prefer the second method
each difference (+2, +3, +4, +5, ...) is separated by 1 each time
this second level of finite differences suggests we have a 2nd degree polynomial

- jim_thompson5910

|dw:1443916308731:dw|

- jim_thompson5910

the general quadratic is ax^2 + bx + c

- jim_thompson5910

plug in x = 1 and we get
ax^2 + bx + c = a*1^2 + b*1 + c = 1a+1b+c = a+b+c
the input is 1, and the output is 1
|dw:1443916381381:dw|
so a+b+c = 1

- jim_thompson5910

next plug in x = 2 (circled) to get
ax^2 + bx + c
a*2^2 + b*2 + c
4a+2b+c
this is equal to the output of 3 (circled) in the second row
|dw:1443916438115:dw|
so 4a+2b+c = 3

- jim_thompson5910

|dw:1443916456906:dw|
plug in x = 3
ax^2 + bx + c
a*3^2 + b*3 + c
9a+3b+c
the output is 6, so
9a+3b+c = 6

- jim_thompson5910

we now have this system of equations
a+b+c = 1
4a+2b+c = 3
9a+3b+c = 6

- Karon86

I am still here

- jim_thompson5910

do you know how to solve for a,b,c?

- jim_thompson5910

you have a number of options
1) elimination
2) substitution
3) matrices

- Karon86

Dude. Like, That is totally different from what I am taught by the book

- amistre64

i contend that it can be any value we want it to be if there are no other conditions for the sequence.

##### 1 Attachment

- jim_thompson5910

@Karon86 how does the book show you? what method do they use?

- Karon86

To find the explicit formal, expand the first few terms of the sequence(sorry I have a terrible iPad):
A1: 1
A2: 1+2 = 3

- Karon86

A3: 1+2+3 = 6
A4: 1+2+3+4=10

- Karon86

A5:1+2+3+4+5=15
An: 1+2+...+n

- Karon86

Are you still here

- jim_thompson5910

I see, let me think

- jim_thompson5910

can you find an explicit formula for An: 1+2+...+n

- Karon86

I don't know, the book doesn't says:
Therefore:
An=1+2+3+...+(n-2)+(n-1)+n. (this is whatI I am so confused)
Which you can write as
An=n+(n-1)+(n-2)+....+3+2+1

- Karon86

Why is it that way?

- Karon86

I am confused with what comes after therefore

- jim_thompson5910

notice how the n pairs with 1, n-1 pairs with 2, n-3 with 3, etc etc

- jim_thompson5910

if you were to add up those pairs you'd get
n+1 = n+1
n-1+2 = n+1
n-2+3 = n+1
...
...
...
3+n-2 = n+1
2+n-1 = n+1
1+n = n+1

- Karon86

I don't understand

- jim_thompson5910

this video might help make things click a bit better?
https://www.youtube.com/watch?v=aaFrAFZATKU
let me know if it doesn't

- Karon86

I cannot acces the YouTube video because it won't let me on th iPad. Please forgive me. Could you tell me the name of the YouTube video

- jim_thompson5910

do you have access to a laptop or desktop computer?

- Karon86

No sir but if I can acces it on my phone. God I wish I had a laptop of desktop

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