Question

0.3,8,15, 24,35,___. numberseries
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Answer 1

Answer:

One possible pattern is this: [(n^2) - 1] for n being a prime power, that is, a positive integer power of a single prime number.

Prime powers are those positive integers that are divisible by exactly one prime number, knowing that number 1 is not counted as a prime power. Prime powers form the sequence as follows: 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17…

There is no (6^2) - 1 = 35, because 6 is not a prime power (6=2x3, 2 and 3 are prime numbers).

So, we have:

(2^2) - 1 = 3

(3^2) - 1 = 8

Answer 2

Answer:

I've written your numbers in the top row of the following array

I've written your numbers in the top row of the following array03385201572024920351120481320

I've written your numbers in the top row of the following array03385201572024920351120481320 Each of the rows below is formed by the difference of the elements above it (directly above minus the one to the left)

I've written your numbers in the top row of the following array03385201572024920351120481320 Each of the rows below is formed by the difference of the elements above it (directly above minus the one to the left)Notice that the third row stabilizes at 2 and the following (and all other) rows are 0. This gives us confidence to believe that this system is giving a reliable prediction. So we extend the second row to 13 and produce 48 as the first value not provided by your question and it should be clear how to proceed from here.

Step-by-step explanation:

Without using this apparatus we still know the following:

Without using this apparatus we still know the following:The third row is represented by the polynomial

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + c

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0f_3(1) = b + c = 3

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0f_3(1) = b + c = 3f_3(2) = 2a + 2b + c = 8

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0f_3(1) = b + c = 3f_3(2) = 2a + 2b + c = 8From this information we can easily recover a, b, and get f_3(n) explicitly.

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0f_3(1) = b + c = 3f_3(2) = 2a + 2b + c = 8From this information we can easily recover a, b, and get f_3(n) explicitly.It's useful in problems of this kind to use base polynomials of the form

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0f_3(1) = b + c = 3f_3(2) = 2a + 2b + c = 8From this information we can easily recover a, b, and get f_3(n) explicitly.It's useful in problems of this kind to use base polynomials of the formn(n-1)(n-2)

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0f_3(1) = b + c = 3f_3(2) = 2a + 2b + c = 8From this information we can easily recover a, b, and get f_3(n) explicitly.It's useful in problems of this kind to use base polynomials of the formn(n-1)(n-2)etc

Without using this apparatus we still know the following:The third row is represented by the polynomialf_3(n)=2 (degree 0)The second row is like this:f_2(n) = 2n+1f_3(n) = a n (n-1) + b n + cWhere f_3(0)= c = 0f_3(1) = b + c = 3f_3(2) = 2a + 2b + c = 8From this information we can easily recover a, b, and get f_3(n) explicitly.It's useful in problems of this kind to use base polynomials of the formn(n-1)(n-2)etcand later expand them to the usual power format. Stirling numbers can be used to automate this.