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.
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
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:
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.
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.
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