The value of the positve integer `n` for which the quadratic equation `sum_(k=1)^n(x+k-1)(x+k)=10n` has solutions `alpha` and `alpha+1` for some `alpha` is

Options

A. 7
B. 11
C. 17
D. 25

Explanation

To find the value of the positive integer \( n \) for which the quadratic equation

\[ \sum_{k=1}^{n}(x+k-1)(x+k) = 10n \]

has solutions \( \alpha \) and \( \alpha + 1 \) for some \( \alpha \), we will follow these steps:

### Step 1: Expand the summation

We start with the left-hand side:

\[ \sum_{k=1}^{n}(x+k-1)(x+k) \]

Expanding the product:

\[ (x+k-1)(x+k) = x^2 + kx + (k-1)x + k(k-1) = x^2 + (2k-1)x + k^2 - k \]

Thus, we have:

\[ \sum_{k=1}^{n}(x+k-1)(x+k) = \sum_{k=1}^{n} \left( x^2 + (2k-1)x + (k^2 - k) \right) \]

### Step 2: Separate the summations

This can be separated into three parts:

\[ \sum_{k=1}^{n} x^2 + \sum_{k=1}^{n} (2k-1)x + \sum_{k=1}^{n} (k^2 - k) \]

Calculating each part:

1. \( \sum_{k=1}^{n} x^2 = n x^2 \)
2. \( \sum_{k=1}^{n} (2k-1) = 2\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 2\frac{n(n+1)}{2} - n = n^2 \)
3. \( \sum_{k=1}^{n} (k^2 - k) = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \)

### Step 3: Combine the results

Putting it all together, we have:

\[ n x^2 + n^2 x + \left( \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \right) = 10n \]

### Step 4: Simplify the equation

The equation can be simplified to:

\[ n x^2 + n^2 x + \left( \frac{n(n+1)(2n+1) - 3n(n+1)}{6} \right) = 10n \]

This simplifies to:

\[ n x^2 + n^2 x + \frac{n(n+1)(2n - 2)}{6} = 10n \]

### Step 5: Rearranging into standard form

Rearranging gives us:

\[ n x^2 + n^2 x + \frac{n(n+1)(2n - 2)}{6} - 10n = 0 \]

### Step 6: Using the properties of roots

Since the roots are \( \alpha \) and \( \alpha + 1 \), we know:

1. The sum of the roots \( \alpha + (\alpha + 1) = -\frac{b}{a} = -\frac{n^2}{n} = -n \)
2. The product of the roots \( \alpha(\alpha + 1) = \frac{c}{a} = \frac{\frac{n(n+1)(2n - 2)}{6} - 10n}{n} \)

### Step 7: Set up equations

From the sum of the roots:

\[ 2\alpha + 1 = -n \implies 2\alpha = -n - 1 \implies \alpha = -\frac{n + 1}{2} \]

From the product of the roots:

\[ \alpha(\alpha + 1) = \frac{n(n+1)(2n - 2) - 60n}{6n} \]

### Step 8: Solve for \( n \)

Setting the equations and simplifying leads us to:

\[ n^2 - 121 = 0 \implies n^2 = 121 \implies n = 11 \]

### Final Answer

Thus, the value of the positive integer \( n \) is:

\[ \boxed{11} \]