What is the formula because that the nth hatchet in the succession 1,4,10,19,31.. Lets say I want to proof because that the 5th term mathematically yet the formula 1+3n, or 1+3(n-1) is not functioning for me, no one is an^2 + bn + c, likewise tried m + n formula and tried to do my own (n-1) formulas with no success or have yet to view a satisfactory simular example.
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Another way to uncover this succession is come again identify the repeated enhancement of multiples that 3.

1=1+3(0)4=1+3(1)10=1+3(3)19=1+3(6)31=1+3(10)

The number in the parantheses after the first term type another popular sequence, the triangle numbers. They are the sum of the an initial n numbers: 1=1,3=1+2,6=1+2+3,10=1+2+3+4 etc...

The...


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Another method to find this succession is come again acknowledge the repeated addition of multiples that 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 an initial term form another popular sequence, the triangular numbers. They room the amount of the an initial n numbers: 1=1,3=1+2,6=1+2+3,10=1+2+3+4 etc...

The `n^(th)` triangular number is discovered 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 around assigning the an initial number in the succession the suitable number; below we begin with n=0.


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

Given the sequence 1,4,10,19,31,... We have the right to view this as a role with intake n and also output f(n). Climate we have actually (1,1),(2,4),(3,10),(4,19),(5,31),...

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

Look at the distinctions of the distinctions (The 2nd order differences)

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

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

The duty will it is in of the kind `y=ax^2+bx+c` and also we are trying to find a,b, and c.

Plug in three well-known x,y pairs to develop three equations in 3 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 because that the succession is:

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

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


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given the sequence:

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

We require to discover the nth formula.

Let us determine the difference between each continuous terms.

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(4-1)= 3

(10-4)= 6

(19-10)= 9

31-19= 12

...

We notification that the difference between each consecutive term deserve to be composed as the formula 3n such that (n= 1,2,3,...)