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You are watching: 1, 4, 10, 19, 31,
Another means to uncover this succession is to again identify the repeated enhancement of multiples that 3.
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...
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Another way to find this succession is to again identify the repeated enhancement of multiples that 3.
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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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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We notification that the difference in between each consecutive term can be composed as the formula 3n such the (n= 1,2,3,...)