What is the formula for the nth hatchet in the succession 1,4,10,19,31.. Lets say I want to proof for the 5th term mathematically however the formula 1+3n, or 1+3(n-1) is not working for me, no one is an^2 + bn + c, also tried m + n formula and tried to make my very own (n-1) formulas through no success or have yet to check out a satisfactory simular example.
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Another means to uncover this succession is to 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 numbers in the parantheses ~ the first term type another famous sequence, the triangular numbers. They room the amount of the very first n numbers: 1=1,3=1+2,6=1+2+3,10=1+2+3+4 etc...

The...

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Another way to find this succession is to 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 numbers in the parantheses after the first term kind 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)` triangle number is found by `(n(n+1))/2` .

So a basic 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 begin with n=0.

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Eric Bizzell
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First, realize that for any type of finite variety of starting facets that there are an infinite variety of sequences through the same starting numbers. However:

Given the succession 1,4,10,19,31,... We have the right to view this as a duty with entry n and output f(n). Climate we have actually (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 in ~ the differences of the differences (The second order differences)

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

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

The role will it is in of the kind `y=ax^2+bx+c` and we are searching for a,b, and also c.

Plug in three well-known x,y bag to create 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 device yields a=3/2,b=-3/2, and also c=1, yielding `y=3/2x^2-3/2x+1` .

So a closed type equation for the sequence is:

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

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

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offered the sequence:

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

We need to find the nth formula.

Let us identify the difference in 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 in between each consecutive term can be composed as the formula 3n such the (n= 1,2,3,...)