Summing the terms of a series

Question

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??

Answer 1

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

Answer 2

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 $$\left(\sum_{i=1}^n i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2.$$

Answer 3

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.

Answer 4

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

Can you proceed from here?

Answer 5

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!

Answer 6

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.