Question

What is the difference between the sum of the cubes and that of the sum of the square of first twenty natural numbers?

• A. 21290
• B. 28630
• C. 36340
• D. 41230

Hint:

Use the standard formulas for sum of cubes and sum of squares of the first n natural numbers with n = 20, then subtract.

Answer 1

The Correct Option is D

Step 1: Recall the formulas for these two sums.
For the first n natural numbers, the sum of the cubes is \( \left(\frac{n(n+1)}{2}\right)^2 \), and the sum of the squares is \( \frac{n(n+1)(2n+1)}{6} \).
Here n = 20, since we want the first twenty natural numbers.

Step 2: Work out the sum of the cubes.
\[ \sum_{k=1}^{20} k^3 = \left(\frac{20 \times 21}{2}\right)^2 = (210)^2 = 44100 \]

Step 3: Work out the sum of the squares.
\[ \sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870 \]

Step 4: Subtract to get the difference.
\[ 44100 - 2870 = 41230 \]

Final Answer:
The difference between the sum of the cubes and the sum of the squares of the first twenty natural numbers is 41230. \[ \boxed{41230} \]