Question:easy

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

Show Hint

Use the standard formulas for sum of cubes and sum of squares of the first n natural numbers with n = 20, then subtract.
Updated On: Jul 21, 2026
  • 21290
  • 28630
  • 36340
  • 41230
Show Solution

The Correct Option is D

Solution and Explanation

Instead of computing the two sums separately and subtracting them, we can build one combined formula for the difference and then plug in n = 20 directly.

The sum of cubes of the first n natural numbers is \(\left(\frac{n(n+1)}{2}\right)^2\), and the sum of squares is \(\frac{n(n+1)(2n+1)}{6}\). Subtracting the second from the first and pulling out the common factor n(n+1):

\[ \sum k^3 - \sum k^2 = n(n+1)\left[\frac{n(n+1)}{4} - \frac{2n+1}{6}\right] \]

Taking the terms inside the bracket to a common denominator of 12:

\[ \frac{n(n+1)}{4} - \frac{2n+1}{6} = \frac{3n(n+1) - 2(2n+1)}{12} = \frac{3n^2 - n - 2}{12} \]

The quadratic \(3n^2 - n - 2\) factors as \((3n+2)(n-1)\), so the whole difference simplifies to a single clean formula:

\[ \sum k^3 - \sum k^2 = \frac{n(n+1)(n-1)(3n+2)}{12} \]

Now put n = 20 into this formula:

\[ \frac{20 \times 21 \times 19 \times 62}{12} \]

Multiply the numerator step by step: 20 x 21 = 420, then 420 x 19 = 7980, then 7980 x 62 = 4,94,760. Dividing this by 12 gives 41,230.

Let's summarize:

  • The difference between the sum of cubes and the sum of squares of the first n natural numbers always equals n(n+1)(n-1)(3n+2)/12.
  • Plugging n = 20 into this single formula gives the answer directly, without working out the two sums on their own.

So the required difference is 41230.

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