At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
the jump from 1 to 3 is +2 |dw:1443915905024:dw|
the jump from 3 to 6 is +3 |dw:1443915932403:dw|
the jump from 6 to 10 is +4 |dw:1443915954079:dw|
the jump from 10 to 15 is +5 |dw:1443915976631:dw|
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)
I want an explicit formula
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
the general quadratic is ax^2 + bx + c
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
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
|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
we now have this system of equations a+b+c = 1 4a+2b+c = 3 9a+3b+c = 6
I am still here
do you know how to solve for a,b,c?
you have a number of options 1) elimination 2) substitution 3) matrices
Dude. Like, That is totally different from what I am taught by the book
@Karon86 how does the book show you? what method do they use?
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
A3: 1+2+3 = 6 A4: 1+2+3+4=10
A5:1+2+3+4+5=15 An: 1+2+...+n
Are you still here
I see, let me think
can you find an explicit formula for An: 1+2+...+n
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
Why is it that way?
I am confused with what comes after therefore
notice how the n pairs with 1, n-1 pairs with 2, n-3 with 3, etc etc
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
I don't understand
this video might help make things click a bit better? https://www.youtube.com/watch?v=aaFrAFZATKU let me know if it doesn't
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
do you have access to a laptop or desktop computer?
No sir but if I can acces it on my phone. God I wish I had a laptop of desktop