Find the nth term of a sequence with different differences

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helpme101010

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I need to find the nth term of these 2 sequences and complete it.

1- 1400, 1830, 2560, 3830, ...., ....

2- 2110, 2740, 3830, 5750, ...., ....

These are two different sequences.

Thanks!
 
helpme101010 said:
I need to find the nth term of these 2 sequences and complete it.

1- 1400, 1830, 2560, 3830, ...., ....

2- 2110, 2740, 3830, 5750, ...., ....

These are two different sequences.

Thanks!
Click to expand...
What is the topic of discussion in your class now?

Are yo allowed to use softwares like MS_Excel?
 
The problem is that any finite number of terms could belong to an infinite number of sequences. What you are being asked for here is the simplest formula that will give these terms- and what is "simplest" is a matter of taste.

I would check first for especially simple things like "arithmetic" or "geometric" series but these are not. (You checked "arithmetic" since you say there are "different differnces".) At worst you can use the fact that the exist an nth power polynomial giving any n numbers. These both have four terms so we can fit a cubic polynomial to them.

1) Try \(\displaystyle an^3+ bn^2+ cn+ d\). That has four unknown coefficients so we need four equations. We want p(1)= 1400, p(2)= 1830, p(3)= 2560, and p(3)= 3830. That is
a+ b+ c+ d= 1400
8a+ 4b+ 2c+ d= 1830
27a+ 9b+ 3c+ d= 2560 and
65a+ 16b+ 4c+ d= 3830
Solve those four equations for a, b, and c.

2)
a+ b+ c+ d= 1400
8a+ 4b+ 2c+ d= 1830
27a+ 9b+ 3c+ d= 2560 and
65a+ 16b+ 4c+ d= 3830
Solve those four equations for a, b, and c.
 
HallsofIvy said:
The problem is that any finite number of terms could belong to an infinite number of sequences. What you are being asked for here is the simplest formula that will give these terms- and what is "simplest" is a matter of taste.

I would check first for especially simple things like "arithmetic" or "geometric" series but these are not. (You checked "arithmetic" since you say there are "different differnces".) At worst you can use the fact that the exist an nth power polynomial giving any n numbers. These both have four terms so we can fit a cubic polynomial to them.

1) Try \(\displaystyle an^3+ bn^2+ cn+ d\). That has four unknown coefficients so we need four equations. We want p(1)= 1400, p(2)= 1830, p(3)= 2560, and p(3)= 3830. That is
a+ b+ c+ d= 1400
8a+ 4b+ 2c+ d= 1830
27a+ 9b+ 3c+ d= 2560 and
65a+ 16b+ 4c+ d= 3830
Solve those four equations for a, b, and c.

2)
a+ b+ c+ d= 1400
8a+ 4b+ 2c+ d= 1830
27a+ 9b+ 3c+ d= 2560 and
65a+ 16b+ 4c+ d= 3830
Solve those four equations for a, b, and c.
Click to expand...

Hello Ivy,

Can you give me a hint on how to find a, b and c?.What about d?

Thanks
 
helpme101010 said:
Hello Ivy,

Can you give me a hint on how to find a, b and c?.What about d?

Thanks
Click to expand...
Hi,
Thought I would just add my 2¢. One of the ways to construct a polynomial going through a set of points is to use a Lagrange interpolating polynomial, see
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html

Another is to just construct the polynomial one point at a time from the points given, using simple 'point slope type functions'. To show how the method works lets look at the points (1,3), (2, 5), (3, 9):
First (1,3)
f(x) = 3 + (x-1) f1(x)
for any function f1. Now add the point (2,5)
f(2) = 3 + (2-1) f1(2) = 3 + f1(2) = 5
So f1(2)=2 or
f1(x) = 2 + (x-2) f2(x)
and
f(x) = 3 + (x-1) { 2 + (x-2) f2(x)}
Now add the point (3,9)
f(3) = 3 + 2 {2 + f2(3)} = 7 + 2 f2(3) = 9
So f2(3)=1 or
f2(x) = 1 + (x-3) f3(x)
and
f(x) = 3 + (x-1) { 2 + (x-2) [1 + (x-3) f3(x)]}
and we have a function which goes through our given points. f3(x) can be any function we want but, to keep f(x) at its simplest, we set it to zero and, simplifying things we have
f(x) = 3 + x (x-1) = x2 - x + 3
 
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Ishuda said:
Hi,
Thought I would just add my 2¢. One of the ways to construct a polynomial going through a set of points is to use a Lagrange interpolating polynomial, see
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html

Another is to just construct the polynomial one point at a time from the points given, using simple 'point slope type functions'. To show how the method works lets look at the points (1,3), (2, 5), (3, 9):
First (1,3)
f(x) = 3 + (x-1) f1(x)
for any function f1. Now add the point (2,5)
f(2) = 3 + (2-1) f1(2) = 3 + f1(2) = 5
So f1(2)=2 or
f1(x) = 2 + (x-2) f2(x)
and
f(x) = 3 + (x-1) { 2 + (x-2) f2(x)}
Now add the point (3,9)
f(3) = 3 + 2 {2 + f2(3)} = 7 + 2 f2(3) = 9
So f2(3)=1 or
f2(x) = 1 + (x-3) f3(x)
and
f(x) = 3 + (x-1) { 2 + (x-2) [1 + (x-3) f3(x)]}
and we have a function which goes through our given points. f3(x) can be any function we want but, to keep f(x) at its simplest, we set it to zero and, simplifying things we have
f(x) = 3 + x (x-1) = x2 - x + 3
Click to expand...
So what is the answer lmao ?
 
helpme101010 said:
Hello Ivy,

Can you give me a hint on how to find a, b and c?.What about d?

Thanks
Click to expand...
Do you not know how to solve simultaneous equation. (And I should have said "solve for a, b, c, and d.")

For
a+ b+ c+ d= 1400
8a+ 4b+ 2c+ d= 1830
27a+ 9b+ 3c+ d= 2560 and
65a+ 16b+ 4c+ d= 3830

an obvious first step is to subtract two equations to eliminate d:
(5) (8a+ 4b+ 2c+ d)- (a+ b+ c+ d)= 7a+ 3b+ c= 1830- 1400= 430
(6) (65a+ 15b+ 4c+ d)- (27a+ 9b+ 3c+ d)= 38a+ 6b+ c= 3830- 2560= 1270

We can get a third equation by subtracting the second equation from the third:
(7) (27a+ 9b+ 3c+ d)- (8a+ 4b+ 2c+ d)= 19a+ 5b+ c= 2560- 1830= 730.

Now subtract the equation (5) from equation (6) to eliminate c:
(9) 31a+ 3b= 840
and subtract equation (5) from equation (7) :
(10) 12a+ 2b= 300

Multiply (9) by 2 to get 62a+ 6b= 1680 and multiply (10) by 3 to get 36a+ 6b= 900.
Subtract those two equations to eliminate b: 26a= 780 so a= 780/26= 30.

Now go back and find b, c, and d.
 
HallsofIvy said:
Do you not know how to solve simultaneous equation. (And I should have said "solve for a, b, c, and d.")

For
a+ b+ c+ d= 1400
8a+ 4b+ 2c+ d= 1830
27a+ 9b+ 3c+ d= 2560 and
65a+ 16b+ 4c+ d= 3830

an obvious first step is to subtract two equations to eliminate d:
(5) (8a+ 4b+ 2c+ d)- (a+ b+ c+ d)= 7a+ 3b+ c= 1830- 1400= 430
(6) (65a+ 15b+ 4c+ d)- (27a+ 9b+ 3c+ d)= 38a+ 6b+ c= 3830- 2560= 1270

We can get a third equation by subtracting the second equation from the third:
(7) (27a+ 9b+ 3c+ d)- (8a+ 4b+ 2c+ d)= 19a+ 5b+ c= 2560- 1830= 730.

Now subtract the equation (5) from equation (6) to eliminate c:
(9) 31a+ 3b= 840
and subtract equation (5) from equation (7) :
(10) 12a+ 2b= 300

Multiply (9) by 2 to get 62a+ 6b= 1680 and multiply (10) by 3 to get 36a+ 6b= 900.
Subtract those two equations to eliminate b: 26a= 780 so a= 780/26= 30.

Now go back and find b, c, and d.
Click to expand...

I managed to find the values of b, c and d:
b=-30
c=310
d=1090
But the values do not fit the last equation, which is 65a+ 16b+ 4c+ d= 3830
I think the 65 should be 64 right?

And in equation (6) it should be 16b not 15b right?

Thanks ;)!
 
Last edited:
OK after doing some corrections, I found out the values of a, b, c and d:
a=40
b=-90
c=420
d=1030

So using the formula \(\displaystyle an^3+ bn^2+ cn+ d\) i get 5880 but the answer should be 6520 :(
 
Last edited:
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