# What is the formula for the nth term in the sequence 1,4,10,19,31.. Lets say I want to proof for the 5th term mathematically but the formula 1+3n, or 1+3(n-1) is not working for me, nor is an^2 +...

What is the formula for the nth term in the sequence 1,4,10,19,31..

Lets say I want to proof for the 5th term mathematically but the formula 1+3n, or 1+3(n-1) is not working for me, nor is

an^2 + bn + c, also tried m + n formula and tried to make my own (n-1) formulas with no success or have yet to see a satisfactory simular example.

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Given the sequence:

1, 4, 10,19, 31, ...

We need to find the nth formula.

Let us determine the difference between each consecutive terms.

(4-1)= 3

(10-4)= 6

(19-10)= 9

31-19= 12

...

We notice that the difference between each consecutive term can be wrote as the formula 3n such that (n= 1,2,3,...)

Then the nth formula could be written as follows:

`==gt (a_(n+1) - a_n) = 3n `

`==gt a_n= a_(n+1) -3n`

Let us check:

a3 = a4 -3(3)

10= 19 -9

10= 10

Let us find the 6th term.

==> a5= a6-3(5)

=> a6= a5 +3(5)

==> a6= 31 +15 = 46

**Then the next consecutive term of the sequence is 46.**

First, realize that for any finite number of starting elements that there are an infinite number of sequences with the same starting numbers. However:

Given the sequence 1,4,10,19,31,... we can view this as a function with input n and output f(n). Then we have (1,1),(2,4),(3,10),(4,19),(5,31),...

Look at the differences in the y values. 4-1=3,10-4=6,19-10=9,31-19=12

Look at the differences of the differences (The second order differences)

6-3=3,9-6=3,12-9=3 which are all the same.

Since the second order differences are the same, you have a quadratic function. You know three (actually more) points, so you can find the function:

The function will be of the form `y=ax^2+bx+c` and we are looking for a,b, and c.

Plug in three known x,y pairs to create three equations in three unknowns:

`1=a(1)^2+b(1)+c`

`4=a(2)^2+b(2)+c`

`10=a(3)^2+b(3)+c`

Solving this system yields a=3/2,b=-3/2, and c=1, yielding `y=3/2x^2-3/2x+1` .

**So a closed form equation for the sequence is:**

`f(n)=1/2(3n^2-3n+2)` where the sequence begins with n=1, and the domain of the function is `n in NN` . (f(1)=1,f(2)=4,etc...)

** Another possible equation is `f(n)=1/2(3n^2+3n+2)` where the sequence begins with n=0. This is sometimes necessary or at least desirable. Here f(0)=1,f(1)=4,f(2)=10 etc... **

Another way to find this sequence is to again recognize the repeated addition of multiples of 3.

1=1+3(0)

4=1+3(1)

10=1+3(3)

19=1+3(6)

31=1+3(10)

The numbers in the parantheses after the first term form another well-known sequence, the triangular numbers. They are the sum of the first n numbers: 1=1,3=1+2,6=1+2+3,10=1+2+3+4 etc...

The `n^(th)` triangular number is found by `(n(n+1))/2` .

So a general formula is `1+(3(n(n+1)))/2` , or `2/2+(3n^2+3n)/2=1/2(3n^2+3n+2)` . Again you have to be careful about assigning the first number in the sequence the proper number; here we start with n=0.