`7^(2)+8^(2)+9^(2)+...+20^(2)=`

Options

A. 2779
B. 7279
C. 7729
D. none of these

Answer

To solve the problem \( 7^2 + 8^2 + 9^2 + \ldots + 20^2 \), we can use the formula for the sum of squares of the first \( n \) natural numbers. The formula is:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 1: Calculate the sum of squares from 1 to 20
We first calculate the sum of squares from \( 1^2 \) to \( 20^2 \):

\[ \sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 \cdot 20 + 1)}{6} \]

Calculating this step-by-step:

- \( n = 20 \)
- \( n + 1 = 21 \)
- \( 2n + 1 = 41 \)

Now substituting these values into the formula:

\[ \sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} \]

### Step 2: Simplify the calculation
Now we simplify \( \frac{20 \cdot 21 \cdot 41}{6} \):

1. Calculate \( 20 \cdot 21 = 420 \)
2. Now, multiply \( 420 \cdot 41 \):

\[ 420 \cdot 41 = 17220 \]

3. Divide by 6:

\[ \frac{17220}{6} = 2870 \]

So,

\[ \sum_{k=1}^{20} k^2 = 2870 \]

### Step 3: Calculate the sum of squares from 1 to 6
Next, we calculate the sum of squares from \( 1^2 \) to \( 6^2 \):

\[ \sum_{k=1}^{6} k^2 = \frac{6(6+1)(2 \cdot 6 + 1)}{6} \]

Calculating this step-by-step:

- \( n = 6 \)
- \( n + 1 = 7 \)
- \( 2n + 1 = 13 \)

Substituting these values into the formula:

\[ \sum_{k=1}^{6} k^2 = \frac{6 \cdot 7 \cdot 13}{6} \]

### Step 4: Simplify the calculation
Now we simplify \( \frac{6 \cdot 7 \cdot 13}{6} \):

1. The \( 6 \) in the numerator and denominator cancels out:

\[ 7 \cdot 13 = 91 \]

So,

\[ \sum_{k=1}^{6} k^2 = 91 \]

### Step 5: Calculate the sum from 7 to 20
Now we find the sum from \( 7^2 \) to \( 20^2 \):

\[ \sum_{k=7}^{20} k^2 = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{6} k^2 \]

Substituting the values we calculated:

\[ \sum_{k=7}^{20} k^2 = 2870 - 91 = 2779 \]

### Final Answer
Thus, the value of \( 7^2 + 8^2 + 9^2 + \ldots + 20^2 \) is:

\[ \boxed{2779} \]