Question

For a positive integer $n$, the quadratic equation $x(x+1)+(x+1)(x+2)+......+(x+n-1)(x+n)=10n$ 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 $11$

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

$x^2+x+x^2+(1+2)x+1.2+x^2+(2+3)x+2.3+....+x^2+(n-1+n)x+(n-1).n=10n$

$n{x}^2+[1+1+2+2+3+3+4+4+....+(n-1)+(n-1)+n]x+[1.2+2.3+3.4+....+(n-1).n]=10n$

Let $1.2+2.3+3.4+...+(n-1).n=S$

$n{x}^2+[2\left(\sum _{r=1}^{n-1}r\right)+n]x+S=10n$

$n{x}^2+(2\frac{(n-1)n}{2}+n)x+S=10n$

$n{x}^2+n^{2}x+S=10n$

Now, $S=\sum _{r=1}^{n-1}r(r+1)$

$S=\sum _{r=1}^{n-1}(r^2+r)$

$S=\frac{(n-1)(n-1+1)(2n-2+1)}{6}+\frac{(n-1)(n-1+1)}{2}$

$S=\frac{n(n-1)(2n-1)}{6}+\frac{n(n-1)}{2}$

$S=\frac{n(n-1)}{2}[\frac{2n-1}{3}+1]$

$S=\frac{n(n-1)}{2}\left[\frac{2n+2}{3}\right]$

$S=\frac{n(n-1)(n+1)}{3}$

Therefore, the equation becomes

$n{x}^2+n^{2}x+\frac{n(n-1)(n+1)}{3}=10n$

$3n{x}^2+3n^{2}x+n^3-n=30n$

$3n{x}^2+3n^{2}x+n^3-31n=0$

For integral solutions of $x$, the discriminant of the equation should be a perfect square.

$\Delta =9n^4-12n(n^3-31n)$

$\Delta =372n^2-3n^4$

$\Delta =3n^2(124-n^2)$

Now, check the options one by one.

A. $n=12$

$\Delta =3\left(144\right)(124-144)<0$. Hence, option $A$ cannot be the answer.

B. $n=9$

$\Delta =3\left(9^2\right)(124-81)=129\left(9^2\right)$ which is not a perfect square. Hence, option $B$ cannot be the answer.

C. $n=10$

$\Delta =3\left({10}^2\right)(124-100)=72\left({10}^2\right)$ which is not a perfect square. Hence, option $C$ cannot be the answer.

D. $n=11$

$\Delta =3\left({11}^2\right)(124-121)=9\left({11}^2\right)$ which is a perfect square. Hence, option $D$ is the only possible answer.