Find `1^(2) + 2^(2) +.................+ 10^(2)`

Explanation

To find the sum of the squares of the first 10 natural numbers, we can use the formula for the sum of squares of the first \( n \) natural numbers:

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

Here, \( n = 10 \).

Let's calculate step by step:

### Step 1: Identify the values
\[ n = 10 \]

### Step 2: Substitute \( n \) into the formula
\[ S = \frac{10(10+1)(2 \cdot 10 + 1)}{6} \]

### Step 3: Simplify inside the parentheses
\[ 10 + 1 = 11 \]
\[ 2 \cdot 10 + 1 = 21 \]

### Step 4: Substitute these values back into the formula
\[ S = \frac{10 \cdot 11 \cdot 21}{6} \]

### Step 5: Perform the multiplication
\[ 10 \cdot 11 = 110 \]
\[ 110 \cdot 21 = 2310 \]

### Step 6: Divide by 6
\[ S = \frac{2310}{6} = 385 \]

So, the sum of the squares of the first 10 natural numbers is \( 385 \).