Question

The value of positive integer $n$ for which the quadratic equation, $\sum _{k=1}^n(x+k-1)(x+k)=10n$ has solutions $\alpha$ and $\alpha +1$ for some $\alpha$, is

Answer 1

$\sum _{k=1}^n(x+k-1)(x+k)=10n$
$\Rightarrow \sum _{k=1}^n[x^2+(2k-1)x+(k-1\left)k\right]=10n$
Now, $\sum _{k=1}^n(2k-1)=\frac{2n(n+1)}{2}-n=n^2$
$\sum _{k=1}^{n}k(k-1)=\sum _{k=1}^n{k}^2-k=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}=\frac{n(n^2-1)}{3}$
So, $n{x}^2+n^{2}x+\frac{n(n^2-1)}{3}-10n=0\Rightarrow 3x^2+3nx+(n^2-31)=0\cdots \left(1\right)$

Now, given that the roots of equation $\left(1\right)$ is $\alpha ,\alpha +1$
Sum of roots
$2\alpha +1=-n\Rightarrow \alpha =-\frac{n+1}{2}\cdots \left(2\right)$
Product of roots
$\alpha (\alpha +1)=\frac{n^2-31}{3}$
Using equation $\left(2\right)$,
$-\frac{n+1}{2}(-\frac{n+1}{2}+1)=\frac{n^2-31}{3}\Rightarrow \frac{n^2-1}{4}=\frac{n^2-31}{3}\Rightarrow n^2=121\therefore n=11$