Thread: Quadratic in Recurrence relation

Hi, first time poster.

Q. In the sequence u1, u2, u3,…,un,
u1 = 0, u2 = 3, u3 = 12 and un = a + bn + cn2
Find the values of a, b and c.

Attempt:
If n = 1 then u1 = a + b(1) + c(1)2 = 0
= a + b + c = 0
If n = 2 then u2 = a + b(2) + c(2)2 = 3
= a + 2b + 4c = 3
If n = 3 then u3 = a + b(3) + c(3)2 = 12
= a + 3b + 9c = 12

= a + b + c = 0
= a + 2b + 4c = 3
= a + 3b + 9c = 12

It's been a number of years since I've last studied maths. Can anyone help me on what comes next? Many thanks.

a + b + c = 0
a + 2b + 4c = 3
a + 3b + 9c = 12

subtract 1st equation from the 2nd ...

b + 3c = 3

subtract 2nd equation from the 3rd ...

b + 5c = 9

solve for b and c using these two equations

Are you able to use the linear combination method to solve systems?

There's also one mistake, the last equation has to be:

Originally Posted by GrigOrig99

Hi, first time poster.

Q. In the sequence u1, u2, u3,…,un,
u1 = 0, u2 = 3, u3 = 12 and un = a + bn + cn^2
Find the values of a, b and c.

Attempt:
If n = 1 then u1 = a + b(1) + c(1)2 = 0
= a + b + c = 0
If n = 2 then u2 = a + b(2) + c(2)2 = 3
= a + 2b + 4c = 3
If n = 3 then u3 = a + b(3) + c(3)2 = 12
= a + 3b + 9c = 12

= a + b + c = 0
= a + 2b + 4c = 3
= a + 3b + 9c = 12

It's been a number of years since I've last studied maths. Can anyone help me on what comes next? Many thanks.

a+b+c=0 gives a+2b+4c=(a+b+c)+b+3c=3=0+3
a+2b+4c=3 gives a+3b+9c=(a+2b+4c)+b+5c=12=3+9

Then b+3c=3 gives b+5c=(b+3c)+2c=9=3+6

Originally Posted by Siron

Are you able to use the linear combination method to solve systems?

There's also one mistake, the last equation has to be:

the OP meant ... the original system is correct.

Fantastic! I've just managed to work it out. Thanks for all the help guys. Awesome site.