Question

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

• A. $9$
• B. $10$
• C. $1$
• D. $12$

Answer 1

The correct option is C $1$

$x(x+1)+(x+1)(x+2)+......+(x+n-1)(x+n)=10n$ -Equation 1

Let the two consecutive integral solutions to this be $\alpha ,\alpha +1$

From Equation 1,$(x^2+x)+(x^2+3x+2)+(x^2+5x+6)+......+[x^2+(2n-1)x+n(n-1\left)\right]=10n$ - Equation 1

For the above quadratic equation, $a=n$

$b=1+1+2+2+3+3+4+4+5+......+(n-1)+(n-1)+n$

$=2(1+2+3+.....+n-1)+n$

$=(n-1)n+n$

$=n^2$ - Equation 2

$c=0\times 1+1\times 2+2\times 3+3\times 4+.....+(n-1)\times n$

$c=\sum (n^2-n)$

$c=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}=n(n-1)(n+1)$ - Equation 3

Sum of roots$=d+d+1=\frac{-b}{a}=\frac{-n^2}{n}=-n$

$\Rightarrow 2d+1=-n$ -Equation 4

Product of roots$=(d^2+d)=n^2-1$ -Equation 5

From Equation 4 & 5, $n^2+2n+1-2n-2=4n^2-4$

$\Rightarrow 3n^2-3=0$

$\Rightarrow n^2-1=0$

$\Rightarrow n\ne -1,n=1$