# Summing the terms of a series [closed]

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

Find the value of:

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

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

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Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: we want to see that you have put significant work into the problem. For example, have you tried to come up with a formula for adding the first $n$ cubic numbers? – Rory Daulton Feb 27 at 16:54

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

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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