1. Key Concepts and Formulas Used

• Summation Notation ($\sum$): Represents a series of terms compactly.
• Sum of the First n Odd Natural Numbers: $\sum_{k=1}^n (2k-1) = n^2$.
• Sum of Products of Consecutive Integers: $\sum_{k=1}^n k(k+1) = \frac{n(n+1)(n+2)}{3}$.
• Vieta's Formulas: For $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$.

2. Step-by-Step Derivation

Step 1: Express the given series in summation form.

We are given the equation: $x(x+1) + (x+1)(x+2) + \dots + (x+n-1)(x+n) = 10n$ We can rewrite this using summation notation: $\sum_{r=1}^n (x+r-1)(x+r) = 10n$ Explanation: The summation represents the sum of $n$ terms. Each term is the product of $(x+r-1)$ and $(x+r)$, where $r$ ranges from $1$ to $n$.

Step 2: Expand the general term of the sum.

Expand the expression $(x+r-1)(x+r)$: $(x+r-1)(x+r) = x^2 + xr + xr + r^2 - x - r = x^2 + (2r-1)x + r(r-1)$ Explanation: We expand the product to obtain a quadratic expression in $x$, which is easier to handle within the summation.

Step 3: Apply summation rules to each term.

Substitute the expanded form back into the summation: $\sum_{r=1}^n (x^2 + (2r-1)x + r(r-1)) = 10n$ Split the summation into three separate summations: $\sum_{r=1}^n x^2 + \sum_{r=1}^n (2r-1)x + \sum_{r=1}^n r(r-1) = 10n$ Evaluate each part:

• First Term: $\sum_{r=1}^n x^2 = nx^2$ Explanation: Since $x^2$ is independent of $r$, it is summed $n$ times.
• Second Term: $\sum_{r=1}^n (2r-1)x = x \sum_{r=1}^n (2r-1) = xn^2$ Explanation: We factor out $x$ and use the identity $\sum_{r=1}^n (2r-1) = n^2$.
• Third Term: $\sum_{r=1}^n r(r-1) = \sum_{r=1}^n (r^2 - r) = \sum_{r=1}^n r^2 - \sum_{r=1}^n r$. We can also write $\sum_{r=1}^n r(r-1) = \sum_{r=1}^n (r^2 - r) = \sum_{r=1}^n r(r+1-2)=\sum_{r=1}^n r(r+1)-2\sum_{r=1}^n r$. Alternatively, we can write $\sum_{r=1}^n r(r-1) = \sum_{r=0}^{n-1} r(r+1) = \frac{(n-1)(n)(n+1)}{3} = \frac{n(n^2-1)}{3}$. Explanation: Using the formula $\sum_{k=1}^n k(k+1) = \frac{n(n+1)(n+2)}{3}$. Here, $N=n-1$, so $\sum_{r=1}^n r(r-1) = \frac{(n-1)n(n+1)}{3}$.

Step 4: Formulate the quadratic equation.

Substitute the evaluated sums back into the original equation: $nx^2 + n^2x + \frac{n(n^2-1)}{3} = 10n$ Divide the entire equation by $n$ (since $n$ is a positive integer, $n \neq 0$): $x^2 + nx + \frac{n^2-1}{3} = 10$ Rearrange the equation to get the standard quadratic form: $x^2 + nx + \frac{n^2-1}{3} - 10 = 0$ $x^2 + nx + \frac{n^2-31}{3} = 0$

Step 5: Apply Vieta's formulas for consecutive integral solutions.

Let the consecutive integral solutions be $\alpha$ and $\alpha+1$. From Vieta's formulas:

• Sum of roots: $\alpha + (\alpha+1) = -n \implies 2\alpha + 1 = -n \implies \alpha = \frac{-n-1}{2}$
• Product of roots: $\alpha(\alpha+1) = \frac{n^2-31}{3}$

Step 6: Solve for n.

Substitute $\alpha = \frac{-n-1}{2}$ into the product of roots equation: $\left(\frac{-n-1}{2}\right)\left(\frac{-n-1}{2} + 1\right) = \frac{n^2-31}{3}$ $\left(\frac{-n-1}{2}\right)\left(\frac{-n+1}{2}\right) = \frac{n^2-31}{3}$ $\frac{(n+1)(n-1)}{4} = \frac{n^2-31}{3}$ $\frac{n^2-1}{4} = \frac{n^2-31}{3}$ $3(n^2-1) = 4(n^2-31)$ $3n^2 - 3 = 4n^2 - 124$ $n^2 = 121$ $n = \pm 11$ Since $n$ is a positive integer, $n = 11$.

3. Tips and Common Mistakes to Avoid

• Careless Expansion: Be careful when expanding and simplifying the expressions. A small mistake can lead to a wrong answer.
• Sign Errors: Ensure correct signs are used during the application of Vieta's formulas.
• Conditions on Variables: Remember that $n$ is a positive integer, which eliminates the negative solution.

4. Summary

We expressed the given series in summation form, expanded the terms, and applied summation rules to obtain a quadratic equation. Then, we used Vieta's formulas and the condition of consecutive integral roots to find the value of $n$. The final calculated value for $n$ is $11$.

1. Final Answer

The final answer is \boxed{11}, which corresponds to option (C).