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. 39630
• E. 41230

Hint:

Use \(\sum k^3=[n(n+1)/2]^2\) and \(\sum k^2=n(n+1)(2n+1)/6\) with \(n=20\), then subtract.

Answer 1

Answer: (E)

Step 1: Recall the standard formulas.
Sum of cubes of first \(n\) natural numbers:
$$\sum k^3=\left[\frac{n(n+1)}{2}\right]^2$$
Sum of squares of first \(n\) natural numbers:
$$\sum k^2=\frac{n(n+1)(2n+1)}{6}$$

Step 2: Substitute \(n=20\) into the sum of cubes formula.
$$\frac{20\times21}{2}=210$$
$$\sum k^3=210^2=44,100$$

Step 3: Substitute \(n=20\) into the sum of squares formula.
$$\sum k^2=\frac{20\times21\times41}{6}$$
\(20\times21=420\), and \(420\times41=17,220\). Dividing by 6:
$$\sum k^2=\frac{17,220}{6}=2,870$$

Step 4: Find the difference.
$$44,100-2,870=41,230$$
So the required difference is 41,230. The correct option is (e) 41230.