I was attempting the recent May/June 2023, CAIE past paper for Further Mathematics variant 12. The question states:
2 The cubic equation $x^3+4x^2+6x+1=0$ has roots $\alpha$, $\beta$, $\gamma$.
(a) Find the value of $\alpha^2+\beta^2+\gamma^2$.
(b) Use standard results from the list of formulae (MF19) to show that $$\sum^n_{r=1} \big((\alpha + r)^2+(\beta+r)^2+(\gamma+r)^2\big)=n(n^2+an+b),$$ where $a$ and $b$ are constants to be determined.
For the (a) part, my working is correct as follows:
$$\big( \sum \alpha \big)^2 = \sum \alpha^2 +2\sum\alpha \beta$$ $$16=\alpha^2-2(6)$$ $$\alpha^2+\beta^2+\gamma^2=4$$
For the (b) part, my working is incorrect according to marking scheme. Realizing that $(\alpha+r)^2=\alpha^2+2\alpha+r^2$ and using above information, my working is: $$\sum^n_{r=1} \big((\alpha + r)^2+(\beta+r)^2+(\gamma+r)^2\big)=\sum^n_{r=1} 3r^2-16r+4$$ $$= \frac{1}{2}n(n+1)(2n+1)-8n(n+1)+4n$$ $$=n\bigg(\big(\frac{n}{2}+\frac{1}{2}\big)(2n+1)-8n-4\bigg)$$ $$=n\bigg(n^2+\frac{n}{2}+n+\frac{1}{2}-8n-4\bigg)$$ $$=n\bigg(n^2 - \frac{15}{2}n-\frac{7}{2}\bigg)$$
Is my attempt correct for the part (b)? The marking scheme has the solution in section 2(b). Am I making a mistake or the marking scheme is wrong? I am losing where how they found the sum of $2r(-4)$, where (MF19) states $\sum^n_{r=1}r=\frac{1}{2}n(n+1)$. It is unclear even with guidance how they got it.
Whatever you want to know follows from $$ (x-\alpha)(x-\beta)(x-\gamma)=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma, $$
$$ (\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\alpha\gamma+\beta\gamma), $$ and $$ (\alpha+r)^2+(\beta+r)^2+(\gamma+r)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha+\beta+\gamma)r+3r^2. $$ Well, you'll also need $$ \sum_{r=1}^nr=\frac{n(n+1)}{2} $$ and $$ \sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}{6}. $$