Question

If, for a positive integer $n,$ the quadratic equation, $x(x+1)+(x+1)(x+2)+\cdots +(x+n-1)(x+n)=10n$ has two consecutive integral solutions, then $n$ is equal to:

• A. $12$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is D $11$

$x(x+1)+(x+1)(x+2)+\cdots +(x+n-1)(x+n)=10n$
$\Rightarrow n{x}^2+(1+3+5+\cdots +(2n-1\left)\right)x+1\cdot 2+2\cdot 3+\cdots +(n-1)n-10n=0$
$\Rightarrow n{x}^2+n^{2}x+\frac{n(n^2-1)}{3}-10n=0$
$\Rightarrow x^2+nx+\frac{n^2-31}{3}=0$
Let $r$ and $r+1$ be the roots of the equation.
$\text{Then},2r+1=-n\cdots \left(1\right)$
$\text{and}r(r+1)=\frac{n^2-31}{3}\cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$n^2=121\Rightarrow n=11$