The positive integer n, for which the solutions of the equation x(x+2) + (x+2)(x+4) + + (x…

Question

The positive integer n, for which the solutions of the equation $x(x+2) + (x+2)(x+4) + \cdots + (x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is :

Accepted Answer

The left-hand side contains $n$ terms:

$\sum _{k=0}^{n-1}(x+2k)(x+2k+2)$

Expanding one term,

$(x+2k)(x+2k+2)={x}^{2}+(4k+2)x+4k(k+1)$

Using the standard sums, the equation becomes

$n{x}^{2}+2{n}^{2}x+\frac{4n({n}^{2}-1)}{3}=\frac{8n}{3}$

Since $n$ is positive, divide by $n$:

${x}^{2}+2nx+\frac{4{n}^{2}-12}{3}=0$

The two roots are consecutive even integers, so their difference is $2$. For a quadratic $a{x}^{2}+bx+c=0$, the difference between its roots is

$\frac{\sqrt{D}}{|a|}$

Here $a=1$, so $\sqrt{D}=2$, giving $D=4$.

Now,

$D=(2n{)}^{2}-4(\frac{4{n}^{2}-12}{3})$

$4=4{n}^{2}-\frac{16{n}^{2}-48}{3}$

$4=16-\frac{4{n}^{2}}{3}$

$\frac{4{n}^{2}}{3}=12$

${n}^{2}=9$

As $n$ is positive,

$n=3$

For $n=3$, the roots are $-4$ and $-2$, which are consecutive even integers.

Hence, the correct option is Option A: $3$.

Solutions · 1