Do you see a pattern

Note that the equations can be written as $ax^2 + b(x+1)^2 + c(x+2)^2 = 2x + 1$ with roots of $1,2,3$ .

By comparing coefficients of $x^2$ , we have $a+b+c=0$ .
By comparing coefficients of $x$ , we have $2b + 4c=2$ .
By comparing coefficients of the constant term, we have $b+4c=1$ .

Solving these equations simultaneously gives us $a=-1,b=1, c= 0$ .

Taking $\frac{1}{2}$ (First equation)- (Second Equation)+ $\frac{1}{2}$ (Last Equation) gives $a+b+c=\boxed{0}$

Subtract (1) from(2) and (2) from (3). Subtract the 2 new eq. to give 2*(a+b+c)=0

Well, as for me I did the same with Sir Otto. But you could use the Cramer's Rule as well. Am I right?

Let a+b+c=x 1→x+3b+8c=3 2→4x+5b+12c=5 3→9x+7b+16c=7 Add 1and 3 then divide what yo get by 2 and than subtract by 2 →x=0

It is well-known that the differences between successive squares form a list of successive odd numbers: $4 - 1 = 3 \\ 9 - 4 = 5 \\ 16 - 9 = 7 \\ \vdots$ This shows immediately that we must choose $a = -1$ , $b = 1$ , and $c = 0$ .