How can I express $(1+2+\dots+(k+1))^2$ using a $\sum$ instead? Good morning from México, I am in my first semester of Mathematics and I started proving by induction that:
$$\sum_{i=0}^n i^3 = \left(\sum_{i=0}^n i\right)^2$$
This question has been answered before, three times actually, but not with the approach I am looking for so I am stuck here (I know I can do it with the other approaches like $((n(n+1))/2)^2$, but I want to know if it is possible doing this):
$$\sum_{i=0}^k i^3 +(k+1)^3= \left(\sum_{i=0}^k i\right)^2+...$$
Next to the $+$ I tried $(k+1)^2$ (which obviously does not equalize my equation) but after plugging in values to check, I knew I was wrong. So my question after all: Is it possible to express: $(1+2+3+...+(k+1))^2$ with sigma?
From that point I can finish proving it. Thank you.
 A: $$  \left(\sum_{i=0}^{k+1} i\right)^2=\left(\left(\sum_{i=0}^{k} i\right)+k+1\right)^2=\left(\sum_{i=0}^{k} i\right)^2+2\left(\sum_{i=0}^{k} i\right)(k+1)+(k+1)^2$$
$$=\left(\sum_{i=0}^{k} i\right)^2+k(k+1)^2+(k+1)^2=\left(\sum_{i=0}^{k} i\right)^2+(k+1)^3$$
$$=\left(\sum_{i=0}^{k} i^3 \right)+(k+1)^3= \sum_{i=0}^{k+1} i^3 . $$
Is that what you wanted?
A: The idea is $\sum_{i=0}^{k+1} i^3 = [\sum_{i=0}^{k} i^3] + (k+1)^3$.
So if you assume that $\sum_{i=0}^k i^3 = (\sum_{i=0}^k i)^2$ then
Then you know $\sum_{i=0}^{k+1} i^3 =$
$[\sum_{i=0}^{k} i^3] + (k+1)^3 = (\sum_{i=0} i)^2  + (k+1)^3$
....
ANd now the job before you is to prove that
$(\sum_{i=0}^k i)^2 + (k+1)^3 = (\sum_{i=0}^{k+1} i)^2$
And I'd prove that by noting that
$(\sum_{i=0}^{k+1} i)^2 = ([\sum_{i=0}^k i] + (k+1))^2$
Can you finish?
What if you notice that
$ ([\sum_{i=0}^k i] + (k+1))^2=$
$[\sum_{i=0}^k i]^2 + 2[\sum_{i=0}^k i](k+1) + (k+1)^2$.
Note that what you need to prove is
$ (\sum_{i=0}^{k+1} i)^2= \color{blue}{(\sum_{i=0}^k i)^2} + (k+1)^3$
And what you have is
$(\sum_{i=0}^{k+1} i)^2 = ([\sum_{i=0}^k i] + (k+1))^2=\color{blue}{(\sum_{i=0}^k i)^2} + 2[\sum_{i=0}^k i](k+1) + (k+1)^2$
So what you NOW need to prove is
$(k+1)^3 = 2[\sum_{i=0}^k i](k+1) + (k+1)^2$
Can you do that?
A: You can start by splitting the sum like this:
$$ \left(\sum_{i=0}^{k+1}i\right)^2 = \left(\sum_{i=0}^{k}i\right)^2 + 2(k+1)\left(\sum_{i=0}^{k}i\right) + (k+1)^2 $$
Then you know (or you prove) that $2(k+1)\left(\sum_{i=0}^{k}i\right)=k^3+2k^2+k$, and when you rewrite your equality you get
$$ \sum_{i=0}^k i^3 + (k+1)^3 = \left(\sum_{i=0}^{k+1}i\right)^2 + k^3+3k^2+3k+1 \\= \left(\sum_{i=0}^{k+1}i\right)^2 + 2(k+1)\left(\sum_{i=0}^{k}i\right) +(k+1)^2 .$$
A: If you know calculus you can express this as the square of the integral of the function $f(x)=x$
$$\left(\int_0^x \tau d\tau\right)^2 = ([\tau^2/2]_{0}^{x})^2 = x^4/4$$
and the other side of the equality is the integral of the function $g(x) = x^3$
$$\int_0^x \tau^3 d\tau = [\tau^4/4]_{0}^{x} = x^4/4$$
