If `sum _(r=1)^(n) (2r +1) =440` , then n = ……

Options

A. 20
B. 22
C. 21
D. 19

Explanation

To solve the equation \( \sum_{r=1}^{n} (2r + 1) = 440 \), we can break it down step by step.

### Step 1: Expand the Summation
We can rewrite the summation as:
\[ \sum_{r=1}^{n} (2r + 1) = \sum_{r=1}^{n} 2r + \sum_{r=1}^{n} 1 \]
This separates the summation into two parts.

### Step 2: Calculate Each Part
The first part, \( \sum_{r=1}^{n} 2r \), can be simplified:
\[ \sum_{r=1}^{n} 2r = 2 \sum_{r=1}^{n} r = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]
The second part, \( \sum_{r=1}^{n} 1 \), is simply \( n \) since we are adding 1 for each of the \( n \) terms.

### Step 3: Combine the Results
Now, we can combine both parts:
\[ n(n + 1) + n = 440 \]
This simplifies to:
\[ n(n + 1 + 1) = 440 \quad \Rightarrow \quad n(n + 2) = 440 \]

### Step 4: Solve the Quadratic Equation
We can rewrite the equation as:
\[ n^2 + 2n - 440 = 0 \]
Now, we will use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -440 \).

### Step 5: Calculate the Discriminant
First, calculate the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-440) = 4 + 1760 = 1764 \]

### Step 6: Find the Roots
Now, we can find \( n \):
\[ n = \frac{-2 \pm \sqrt{1764}}{2 \cdot 1} \]
Calculating \( \sqrt{1764} = 42 \):
\[ n = \frac{-2 \pm 42}{2} \]
This gives us two possible solutions:
\[ n = \frac{40}{2} = 20 \quad \text{and} \quad n = \frac{-44}{2} = -22 \]
Since \( n \) must be a positive integer, we have:
\[ n = 20 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{20} \]