Question

The sum of the series \[ 1 + 2^2 + 3^2 + 4^2 + \dots + 100^2 \] is

• A. \( 100^2 + 100 \)
• B. \( 99 \times 2^{100} - 1 \)
• C. \( 99 \times 2^{100} + 1 \)
• D. \( 99 \times 2^{100} \)

Hint:

The sum of squares of the first \( n \) natural numbers follows a specific formula for efficient calculation.

Answer 1

The Correct Option is C

Step 1: Use the formula for the sum of squares.
The sum of the first \( n \) squares is given by: \[ S = \frac{n(n+1)(2n+1)}{6} \] For \( n = 100 \), substitute the value to find the sum.
Step 2: Conclusion.
The sum of squares of the first 100 numbers results in \( 99 \times 2^{100} + 1 \). Final Answer: \[ \boxed{99 \times 2^{100} + 1} \]