3,10,21,36..., i know the next # is 55, i know the formula's n(2n+1) (or 2n^2 +n)..i need A formula that....

Question

relates each number to the previous number....like if i had 2,4,6,8,10..i know that the next number is 12 and i know that the formula is 2n, but i need a recursive formula to show how to get from one number to the next...like this:

recursive: a(sub n)= a(subn-1)+2
in other words, to get to the next number, take the PREVIOUS NUMBER and add 2...

once again, i know the next number is 55 and i know that the formula is n(2n+1) or 2n^2 +n.....


does anyone understand what im trying to say....

im not asking u to do my homework...is this right...?

a(subn)= a(subn-1) + (7+4n)...?

sry im very confused about this..if no one understands it its ok

Answer 1

See my answer for your previous posting of this problem. The final answer is

a_(n+1) = a_n + 3 + 4n

Your expression

a_n= a_(n-1) + (7+4n)

is close, but off by 2n. According to your formula, a_2 = a_1 + (7 + 4*2), but that doesn't work, because it gives a_2 = 3 + 7 + 8 = 18 instead of 10.

So you can rewrite it as

a_n = a_(n-1) + [7 + 4(n-2)]

which simplifies to

a_n = a_(n-1) + (4n - 1)

which is equivalent to my original answer.

See my earlier response for more details. Hope that helps!

Answer 2

it often helps to check the differences between the numbers:
10 - 3 = 7
21 - 10 = 11
36 - 21 = 15

if you still don't see the pattern, take differences again:
11 - 7 = 4
15 - 11 = 4
and you see...

You want a recursive formula. You can use the differences 7, 11, 15, ... which increase linearly in steps of four. Therefore they obey the linear formula 4*n + (a number) and it is easy to see that that number should be 3, because 7 = 4*1 + 3, 11 = 4*2 + 3, etc.

That gives you a_{n+1} = a_n + 4*n + 3.


You can even make a better recursion formula as follows. The difference between numbers a_{n+1} and a_{n+2} is 4 more than the previous difference, that between a_n and a_{n+1}. Therefore,

a_{n+2} = a_{n+1} + (4 + a_{n+1} - a_n)

... = 2 a_{n+1} - a_n + 4

For instance,
21 = 2 * 10 - 3 + 4
36 = 2 * 21 - 10 + 4
55 = 2 * 36 - 21 + 4

That way you have removed all direct references to n.

Answer 3

An=2n^2+n
An+1=2(n+1)^2+(n+1)=2n^2+5n+3=2n^2+n+4n+3=an+4n+3
so
An+1=An+4n+3
It look like there is no real recursive formula in this case.

Answer 4

a(sub 1) = 3
a(sub n) = a(sub n-1) + 7 + (n - 2) 4 = a(sub n-1) + 4n - 1 , n >=2.

Answer 5

We understand,but we're NOT going to do your homework for you!