Findd the sum of the squares of first 35 natural numbers.

Options

A. 14910
B. 15510
C. 14510
D. 16510

Answer

To find the sum of the squares of the first 35 natural numbers, we can use the formula for the sum of squares of the first \( n \) natural numbers:

\[ S_n = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 1: Identify \( n \)
In this case, \( n = 35 \).

### Step 2: Substitute \( n \) into the formula
We substitute \( n \) into the formula:

\[ S_{35} = \frac{35(35 + 1)(2 \cdot 35 + 1)}{6} \]

### Step 3: Simplify the expression
Calculate \( 35 + 1 \) and \( 2 \cdot 35 + 1 \):

\[ 35 + 1 = 36 \]
\[ 2 \cdot 35 + 1 = 70 + 1 = 71 \]

Now substitute these values back into the formula:

\[ S_{35} = \frac{35 \cdot 36 \cdot 71}{6} \]

### Step 4: Calculate the product in the numerator
First, calculate \( 35 \cdot 36 \):

\[ 35 \cdot 36 = 1260 \]

Now multiply this result by 71:

\[ 1260 \cdot 71 \]

To calculate \( 1260 \cdot 71 \), we can break it down:

\[ 1260 \cdot 71 = 1260 \cdot (70 + 1) = 1260 \cdot 70 + 1260 \cdot 1 \]

Calculating \( 1260 \cdot 70 \):

\[ 1260 \cdot 70 = 88200 \]

And \( 1260 \cdot 1 = 1260 \).

So,

\[ 1260 \cdot 71 = 88200 + 1260 = 89460 \]

### Step 5: Divide by 6
Now, we divide \( 89460 \) by \( 6 \):

\[ S_{35} = \frac{89460}{6} = 14910 \]

### Final Answer
Thus, the sum of the squares of the first 35 natural numbers is:

\[ \boxed{14910} \]