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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You are watching: What is next in this series? 1, 4, 10, 19, 31, _

Another way to uncover this succession is come again identify the repeated enhancement of multiples that 3.


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...


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Another method to find this succession is come again acknowledge the repeated addition of multiples that 3.


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,...)