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I have a really confusing question from an investigation. It states-

Find the value of:

$$\sqrt{1^3+2^3+3^3+\ldots+100^3}$$

How would I go about answering this??

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closed as off-topic by Rory Daulton, Daniel W. Farlow, Jeel Shah, N. F. Taussig, ᴡᴏʀᴅs Feb 27 at 23:34

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$$\sum_{r=1}^n r^3=\frac {n^2(n+1)^2}{2^2}=\left(\sum_{r=1}^nr\right)^2\\ \sqrt{\sum_{r=1}^{100}r^3}=\sum_{r=1}^{100}r=\binom{101}2=5050$$ i.e. $$\sqrt{1^3+2^3+3^3+\cdots+100^3}=1+2+3+\cdots+100=5050$$

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Hint: show by induction that $$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}.$$

Note that the right-hand side is just $$\left(\sum_{i=1}^n i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2.$$

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HINT:

Prove by induction that $$1^3+2^3+3^3+\ldots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$$ and use that result in your question.

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$$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^{2}}{4}$$

Can you proceed from here?

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And if you don't know the formula and don't need it exactly,

$\sum_{k=1}^{100} k^3 \approx \int_0^{100} x^3 dx =\frac{100^4}{4} $ so the result is $\sqrt{\frac{100^4}{4}} =\frac{100^2}{2} =5000 $.

If you add in the usual correction of $\frac12 f(n)$, the result is $\sqrt{\frac{100^4}{4}+\frac12 100^3} =\frac{100^2}{2}\sqrt{1+\frac{2}{100}} \approx \frac{100^2}{2}(1+\frac{1}{100}) =5050 $.

Shazam!

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There is a nice visual proof (of concept) of the Nicomachus's theorem: the sum of first cubes is a squared triangular number:

Wikipedia: Nichomacus theorem

So the square root of the total volume of the cubes above is the side of the square below, hence the sum of the first integers.

In this example, you have $225$ "unit cubes", whose root is $15$, i. e. the length of the side of the square below.

It takes some more time to draw $100$ increasing cubes, but the proof works along the same lines.

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