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 \(\sum k^3=[n(n+1)/2]^2\) and \(\sum k^2=n(n+1)(2n+1)/6\) with \(n=20\), then subtract.
Updated On: Jul 20, 2026
  • 21290
  • 28630
  • 36340
  • 39630
  • 41230
Show Solution

The Correct Option is

Solution and Explanation

Instead of computing the two sums separately, combine them into a single term-by-term sum first.
We want \(\sum_{k=1}^{20}(k^3-k^2)=\sum_{k=1}^{20}k^2(k-1)\).
Writing this out: \(0^2\times0+\) actually starting cleanly, note \(k^2(k-1)\) for \(k=1\) gives \(1\times0=0\), so the first real contribution starts at \(k=2\). Rather than summing term by term, it is simpler to keep the two closed-form results and subtract, but let's verify the final number using small-scale sums to build confidence in the formulas, then scale up.
For \(n=3\): sum of cubes \(=1+8+27=36=(3\times4/2)^2=6^2=36\) checks out; sum of squares \(=1+4+9=14=3\times4\times7/6=84/6=14\) checks out. Both formulas are verified on a small case.
Now applying them at \(n=20\): sum of cubes \(=(210)^2=44100\), sum of squares \(=\dfrac{20\times21\times41}{6}=2870\).
Difference \(=44100-2870=41230\).\[\boxed{41230}\]
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