Summing the terms of a series 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??
 A: $$\sum_{r=1}^n r^3=\frac {n^2(n+1)^2}{2^2}=\left(\sum_{r=1}^nr\right)^2\\
\sqrt{\sum_{r=1}^{100}r^3}=\sum_{r=1}^{100}r=\binom{101}2=5050$$
i.e.
$$\sqrt{1^3+2^3+3^3+\cdots+100^3}=1+2+3+\cdots+100=5050$$
A: 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 $$\left(\sum_{i=1}^n i\right)^2 = \left(\frac{n(n+1)}{2}\right)^2.$$
A: 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.
A: $$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^{2}}{4}$$
Can you proceed from here?
A: 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!
A: 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.
