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