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

x(x+2)+(x+2)(x+4)+ ... .+(x+2 n-2)(x+2 n)=(8 n / 3) => sum _ r=1^ n( x+2 r-2)( x+2 r)=frac8 n3 n x^2+2 x sum _r=1^n(2 r-1)+4 sum _r=1^n r(r-1)=(8 n / 3) n x^2+2 x * n^2+frac4 n(n^2-1)3-(8 n / 3)=0 $x^2+2 n x+frac4(n^2-1)3-(8 / 3)=0

The positive integer n for which the solutions of the e... | null (Mathematics)

x(x+2)+(x+2)(x+4)+\ldots .+(x+2 n-2)(x+2 n)=\frac{8 n}{3}\\

\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}}(\mathrm{x}+2 \mathrm{r}-2)(\mathrm{x}+2 \mathrm{r})=\frac{8 \mathrm{n}}{3}\\

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

n x^{2}+2 x \cdot n^{2}+\frac{4 n\left(n^{2}-1\right)}{3}-\frac{8 n}{3}=0\\

x^{2}+2 n x+\frac{4\left(n^{2}-1\right)}{3}-\frac{8}{3}=0<_{\beta}^{\alpha}\\

\because|\alpha-\beta|=2 \Rightarrow \frac{\sqrt{\mathrm{D}}}{|\mathrm{a}|}=2 \Rightarrow \mathrm{D}=4\\

\Rightarrow 4 \mathrm{n}^{2}-4\left(4 \frac{\left(\mathrm{n}^{2}-1\right)}{3}-\frac{8}{3}\right)=4\\

\Rightarrow \mathrm{n}^{2}-\frac{4 \mathrm{n}^{2}}{3}=-3\\

\Rightarrow \mathrm{n}^{2}=9\\

\Rightarrow \mathrm{n}=3

Explanation

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