# A-level Further Maths Series and Sequences question HELP

Question: You are given that
n
(ar + b) = n^2
r=1
for all n where a and b are constants. Determine the value of a and b.

So far I've done:
∑ar + ∑b = n^2
an/2 (n+1) + bn = n^2
n/2(an + a + 2b) = n^2
(an^2 + an) / 2 + bn = n^2

I know the values for a and b are 2 and -1 respectively, but can't prove it.
What should I do next?
Original post by amarooDan
Question: You are given that
n
(ar + b) = n^2
r=1
for all n where a and b are constants. Determine the value of a and b.

So far I've done:
∑ar + ∑b = n^2
an/2 (n+1) + bn = n^2
n/2(an + a + 2b) = n^2
(an^2 + an) / 2 + bn = n^2

I know the values for a and b are 2 and -1 respectively, but can't prove it.
What should I do next?

Equate the linear and quadratic coefficients. The "for all n" is a bit of hint that its the way to go.
(edited 3 months ago)
Original post by amarooDan
Question: You are given that
n
(ar + b) = n^2
r=1
for all n where a and b are constants. Determine the value of a and b.

So far I've done:
∑ar + ∑b = n^2
an/2 (n+1) + bn = n^2
n/2(an + a + 2b) = n^2
(an^2 + an) / 2 + bn = n^2

I know the values for a and b are 2 and -1 respectively, but can't prove it.
What should I do next?

Compare coefficients on each side of your last line. you know that the n^2 term has a coefficient of 1 on the right and the n term has a coefficient of zero on the right. On the left, your coefficient for the n^2 term is a/2, and for the n term it's a/2 - b. With this you can get a=2 pretty easily and just use that to calculate b
Original post by mqb2766
Equate the linear and quadratic coefficients. The "for all n" is a bit of hint that its the way to go.

Thanks for the hint, solved it
did you think about why you can equate the coefficients?
A simpler method is making n the subject of the equation
n= (2b+a)/(2-a) and then substituting values for n where n∈Z+ and then it should form a simultaneous equation. and a and b should easily be found.