The positive integer n$n$, for which the solutions of the equation x(x+2)+(x+2)(x+4)+ldots+(x+2n-2)(x+2n) = frac8n3$x(x+2)+(x+2)(x+4)+\ldots+(x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is :

### Related Formula sum_r=1^n (2r-1) = n^2$$\sum_{r=1}^{n} (2r-1) = n^2$$ sum_r=1^n r(r-1) = fracn(n^2-1)3$$\sum_{r=1}^{n} r(r-1) = \frac{n(n^2-1)}{3}$$ ### Core Logic Rewrite the series in summation form: sum_r=1^n(x+2r-2)(x+2r)=frac8n3$$\sum_{r=1}^{n}(x+2r-2)(x+2r)=\frac{8n}{3}$$ nx^2 + 2xsum_r=1^n(2r-1) + 4sum_r=1^nr(r-1) = frac8n3$$nx^2 + 2x\sum_{r=1}^{n}(2r-1) + 4\sum_{r=1}^{n}r(r-1) = \frac{8n}{3}$$ ### Step 1: Simplify the Equation Substitute the standard summation formulas: nx^2 + 2x(n^2) + frac4n(n^2-1)3 - frac8n3 = 0$$nx^2 + 2x(n^2) + \frac{4n(n^2-1)}{3} - \frac{8n}{3} = 0$$ Divide the entire equation by n$n$: x^2 + 2nx + frac4(n^2-1)3 - frac83 = 0$$x^2 + 2nx + \frac{4(n^2-1)}{3} - \frac{8}{3} = 0$$ Let the roots be alpha$\alpha$ and beta$\beta$. ### Step 2: Apply the Condition for Roots Since the roots are two consecutive even integers, their difference is 2$2$. |alpha - beta| = 2 implies fracsqrtD|a| = 2 implies D = 4$$|\alpha - \beta| = 2 \implies \frac{\sqrt{D}}{|a|} = 2 \implies D = 4$$ (2n)^2 - 4(1)left( frac4(n^2-1)3 - frac83 right) = 4$$(2n)^2 - 4(1)\left( \frac{4(n^2-1)}{3} - \frac{8}{3} \right) = 4$$ 4n^2 - frac16(n^2-1)3 + frac323 = 4$$4n^2 - \frac{16(n^2-1)}{3} + \frac{32}{3} = 4$$ n^2 - frac4n^23 = -3$$n^2 - \frac{4n^2}{3} = -3$$ frac-n^23 = -3 implies n^2 = 9$$\frac{-n^2}{3} = -3 \implies n^2 = 9$$ Since n$n$ is a positive integer, n = 3$n = 3$. ### Pattern Recognition When dealing with equations where roots have a specific difference k$k$, instantly use D = a^2k^2$D = a^2k^2$. Here k=2$k=2$, so D=4a^2$D=4a^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations Class 11 Maths: Sequence and Series