Saturday, 16 November 2019

calculus - How do I derive 1+4+9+cdots+n2=fracn(n+1)(2n+1)6












I am introducing my daughter to calculus/integration by approximating the area under y = f(x*x) by calculating small rectangles below the curve.



This is very intuitive and I think she understands the concept however what I need now is an intuitive way to arrive at n(n+1)(2n+1)6 when I start from 1+4+9++n2.



In other words, just how came the first ancient mathematician up with this formula - what were the first steps leading to this equation? That is what I am interested in, not the actual proof (that would be the second step).


Answer



Same as you can prove sum of n = n(n+1)/2 by



*oooo
**ooo

***oo
****o


you can prove n(n+1)(2n+1)6 by building a box out of 6 pyramids:



enter image description here
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enter preformatted text here




Sorry the diagram is not great (someone can edit if they know how to make a nicer one). If you just build 6 pyramids you can easily make the n x n+1 x 2n+1 box out of it.




  • make 6 pyramids (1 pyramid = 1+22+32+42+... blocks)

  • try to build a box out of them

  • measure the lengths and count how many you used.. that gives you the formula



Using these (glued) enter image description here


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