Solution
$\mathrm{nx}^{2}+\mathrm{x}(2+4+6+\ldots . .+2 \mathrm{n})+(1.3+2.4 \ldots . .+(\mathrm{n}-1)(\mathrm{n}+1))=4 \mathrm{n}$
\[
\begin{aligned}
& \therefore 1.3+2.4+\cdots .+(n-1)(n+1) \\
&=\sum_{r=1}^{n-1} r(r+2)=\sum_{r=1}^{n-1} r^{2}+\sum_{r=1}^{n-1} 2 r=\frac{(n-1)(n)(2 n-1)}{6}+\frac{2(n-1) n}{2}=\frac{n(n-1)(2 n+5)}{6} \\
& \Rightarrow n x^{2}+n(n+1) x+\frac{n(n-1)(2 n+5)}{6}=4 n \Rightarrow x^{2}+(n+1) x+\frac{(n-1)(2 n+5)}{6}=4
\end{aligned}
\]
As roots are integers
D must be a perfect square $\mathrm{D}=(n+1)^{2}-4\left[\frac{(\mathrm{n}-1)(2 \mathrm{n}+5)}{6}-4\right]=16+\frac{13-\mathrm{n}^{2}}{3}$
\[
\begin{aligned}
& =20-\left(\frac{\mathrm{n}^{2}-1}{3}\right)=\mathrm{P}^{2} ; \mathrm{P} \in \mathrm{I} ; \mathrm{P} \in\{1,2,3,4\} \\
& \Rightarrow \mathrm{n}^{2}=61-3 \mathrm{P}^{2} \text { For integer, } \mathrm{P}=2, \mathrm{n}=7 \\
& \mathrm{x}^{2}+8 \mathrm{x}+15=0 \Rightarrow \mathrm{x}=-3,-5 \Rightarrow \alpha=-5, \alpha+2=-3 \therefore \alpha+\mathrm{n}=7-5=2
\end{aligned}
\]
[CHAPTER M06 COMPLEX NUMBER TOPIC T1 ]
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