If the roots of the equation x(x + 2) + (x + 1)(x + 3) + …. + (x + (n – 1)) + (x + n + 1) = 4n are α and α + 2. Then, the value of |2 α + n| is

The correct answer is 3.

Let us analyze the given equation step-by-step:

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

The general term of the series on the left-hand side can be written for $r=1,2,...,n$ as:

$T_r=(x+r-1)(x+r+1)={(x+r)}^2-1=x^2+2rx+r^2-1$

Summing these terms from $r=1$ to $n$, we get:

$\sum _{r=1}^n[x^2+2rx+r^2-1]=4n$

Using the standard summation formulas for the first $n$ natural numbers and their squares:

$n{x}^2+2x·\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}-n=4n$

Divide the entire equation by $n$ (since $n\ne 0$):

$x^2+(n+1)x+\frac{(n+1)(2n+1)}{6}-1=4$

Rearranging the terms, we obtain the quadratic equation in $x$:

$x^2+(n+1)x+\frac{2n^2+3n-29}{6}=0$

Let the roots of this quadratic equation be $\alpha$ and $\alpha +2$. The difference between the roots is:

$\left|\right(\alpha +2)-\alpha |=2$

For a quadratic equation $a{x}^2+bx+c=0$, the difference of roots is given by $\frac{\sqrt{D}}{\left|a\right|}$, where $D=b^2-4ac$ is the discriminant. Since $a=1$, we have:

$D=2^2=4$

Substituting the coefficients into the discriminant formula:

${(n+1)}^2-4\left(1\right)·\frac{2n^2+3n-29}{6}=4$

Multiply the entire equation by 3 to eliminate the fraction:

$3(n^2+2n+1)-2(2n^2+3n-29)=12$

Simplifying the expression:

$3n^2+6n+3-4n^2-6n+58=12$

$-n^2+61=12$

$n^2=49$

Since $n$ must be a positive integer, we have:

$n=7$

Now, using the sum of the roots of the quadratic equation:

$\alpha +(\alpha +2)=-(n+1)$

$2\alpha +2=-(7+1)=-8$

$2\alpha =-10$

We need to find the value of $|2\alpha +n|$:

$|2\alpha +n|=|-10+7|=|-3|=3$