Find the 6th, 7th and 8th terms in the following sequence...

[kmj]

2, 5, 10, 18, 30 ...

I'm not asking for the answer, I just need a hint in figuring out the following sequence. Any help would be appreciated. Thanks!

 

[Ishuda]

2, 5, 10, 18, 30 ...

I'm not asking for the answer, I just need a hint in figuring out the following sequence. Any help would be appreciated. Thanks!

There are many ways (actually an uncountable infinite number of ways) to extend the series. What one might start off with is looking at differences,
(1st) 5-2=3, 10-5=5, 18-10=8, 30-18=12
(2nd) 5-3=2, 8-5=3, 12-8=4
(3rd) 3-2=1, 4-3=1
and arrive at a constant series. Since it is the 3rd order difference which is constant, a third degree polynomial will work for the series. This always works but you might end up with a polynomial whose degree is the number of numbers in the series minus 1, i.e. 1 point is a zeroth order polynomial, 2 points is a linear polynomial, etc.

Another method is to take ratios which could work for a series such as 1, 2, 4, 8, 16, ...

Or just 'recognize the series' such as 2, 3, 5, 7, 11, ... is the prime numbers

HOWEVER, none of these methods or any of the myriad other methods may be what was had in mind when the question was asked. You think maybe what they wanted as the following three terms were \(\displaystyle \pi\), e, and 6.3 ?

 

[pka]

2, 5, 10, 18, 30 ...
I'm not asking for the answer, I just need a hint in figuring out the following sequence. Any help would be appreciated.

This one turned out to be a beast.
\(\displaystyle {a_1} = 2,\quad {a_{n }} = {a_{n - 1}} + \left\lfloor {\dfrac{{{{\left( {n + 1} \right)}^2}}}{3}} \right\rfloor \) that is the floor function.

Look here.

 

[Ishuda]

This one turned out to be a beast.
\(\displaystyle {a_1} = 2,\quad {a_{n }} = {a_{n - 1}} + \left\lfloor {\dfrac{{{{\left( {n + 1} \right)}^2}}}{3}} \right\rfloor \) that is the floor function.

Look here.

Actually a cubic will fit it exactly, see WolframAlpha

 

[pka]

Actually a cubic will fit it exactly, see WolframAlpha

What cubic? Fits what (it)?

Have a look at this.

EDIT: Is this what you mean?

 

[Ishuda]

n(n^2 + 11) / 6 works fine...10th term = 10(111)/6 = 185

Which is the Wolfram Alpha solution

 

[Ishuda]

What cubic? Fits what (it)?

Have a look at this.

EDIT: Is this what you mean?

Yes on the latter link as Denis pointed out:
a_n = n (n² + 11) / 6
a₁ = 1 (1² + 11) / 6 = 2
a₂ = 2 (2² + 11) / 6 = 5
a₃ = 3 (3² + 11) / 6 = 10
a₄ = 4 (4² + 11) / 6 = 18
a₅ = 5 (5² + 11) / 6 = 30
...

However, you could also do
a_n = n (n² + 11) / 6 + (x-1)(x-2)(x-3)(x-4)(x-5) g(n)
for any bounded function g.

 

[Deleted member 4993]

(n² + 11) is an interesting term.

It is divisible by 3 - iff n is NOT divisible by 3.

if n = 3m + 1 → n² + 11 = 9m² + 6m + 12, and

if n = 3m + 2 → n² + 11 = 9m² + 12m + 15

By the way, what is the reason for the assumption that the sequence in question is made of integers only??!!