Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.
 A: Hint: the left hand side equals $$(1+n)^3=(1+n)^2+n(1+n)^2$$
and $$\sum_{k=0}^nk=\frac{n(n+1)}{2}$$
A: The sum $P(n):=1+2+\cdots n$ must be a quadratic polynomial in $n$, because $P(n)-P(n-1)=n$ is a linear polynomial.
Then as both members are cubic polynomials, it suffices to check equality for four distinct values of $n$.
$$\begin{align}n=0&\to 1^3=2()\cdot1+1^2&=1,\\
n=1&\to 2^3=2(1)\cdot2+2^2&=8,\\
n=2&\to 3^3=2(1+2)\cdot3+3^2&=27,\\
n=3&\to 4^3=2(1+2+3)\cdot4+4^2&=64.\end{align}
$$
This proves the identity for all $n$.

With the simplified identity (in my other answer), checking for three values suffices
$$\begin{align}n=0&\to 1^2=2()+1&=1,\\
n=1&\to 2^2=2(1)+2&=4,\\
n=2&\to 3^2=2(1+2)+3&=9.\end{align}
$$
A: You just need to know $1+2+...+n = \frac {n(n+1)}{2}$. Substitute this in and you will see the RHS equals $(1+n)^2+n(1+n)^2$. Factorize it and you will get the answer you need (i.e. the trinomial in the left hand side).

To obtain the first identity, you can pair the first number with the last number in the sequence, then the second with the second to last and so on (i.e. $1+n$ and $2+(n-1)$ and $3+(n-2)$) you can see that these sums will always add up to $n+1$ since you are pairing elements together, and you have $n$ elements in the sum, you will have $n/2$ lots of $n+1$ multiplying these two together yields the first identity. 

There is a nice story accompanying this identity involving a young child named Gauss, I don't know if it is true or not but it is famous nonetheless. The math teacher at his school wanted to discipline the children for being rowdy and made them add all the numbers between one and one hundred. Gauss was over in seconds while the rest of his classmates toiled away. He used (presumably) the above method to get $5050$
A: You can simplify by $n+1$ to get
$$(n+1)^2=2(1+2+\cdots n)+(n+1).$$
Then by induction,
$$(n+1)^2-n^2=2n+1=2(1+2+\cdots n)-2(1+2+\cdots n-1)+(n+1)-n.$$
