Proving $(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3$ for all $n\in\mathbb N$ and all $a\ge -1.$ I was asked to prove the the following without induction. Could someone please verify whether my proof is right? Thank you in advance.

For any real number $a\ge -1$ and every natural number n, the statement, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ holds.

Proof. From the binomial theorem we can see that, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ becomes, $$\sum^{n}_{k=0}\binom{n}{k}a^{k}\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}.$$ Also, if $a\lt b$, let us define $\sum^{a}_{i=b}f(i)=0$, by convention. Then for n=1 we have,\begin{align}\sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-\sum^{1}_{k=4}\binom{n}{k}a^{k}\\ \sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-0,\end{align} which is true, since both remaining sums are equal to each other. It is similarly the case for $n=2$ and $n=3$. More generally, \begin{align}\sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0.\end{align} 

To determine whether this last statement is true, we first consider $a\ge0$, and we see that it is trivially true. Next we consider the case where $-1\le a\lt0.$ We let $a=-b$, so $0\lt b\le 1$ and notice that, $$\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}$$ then becomes, $$\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^{k}\binom{n}{k}b^{k}+...+(-1)^{n}\binom{n}{n}b^{n}.$$
Also from the binomial theorem we can see that, 
  \begin{align}
(1+(-1))^{n}-(1+(-1))^{3}+(1+(-1))^{1}\\
=\sum^{n}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}-\sum^{3}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}+\sum^{1}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}\\ 
0=1-\binom{n}{1}+\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\\ 
n-1=\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\ge0,
\end{align} 
  and by noticing that $b^4\ge b^5\ge b^6\ge ...\ge b^k$, we can see that each binomial term $\binom{n}{k}$ is multiplied by a factor of $b^k$, making each term smaller than the term before. 
Thus,
  \begin{align}
\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^n\binom{n}{n}b^n\ge0\\ 
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{n}a^n\ge0,\tag{by substitution}
\end{align} as desired.
Working backward, \begin{align}\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ (1+a)^n&\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3.\ \ \ \blacksquare\end{align}

 A: This is not an answer to the question about your proof.  I am offering an alternative proof,using Taylor's Theorem.  Note that, for integers $n$ and $k$ with $n> k\geq 0$,
$$(1+a)^n-\sum_{r=0}^k\,\binom{n}{r}\,a^r=(k+1)\binom{n}{k+1}\,\int_{0}^a\,(1+x)^{n-k-1}\,(a-x)^k\,\text{d}x\,.$$
Your question is a particular case where $k=3$, which leads to
$$(1+a)^n-\sum_{r=0}^3\,\binom{n}{r}\,a^r=4\binom{n}{4}\,\int_{0}^a\,(1+x)^{n-4}\,(a-x)^3\,\text{d}x\geq 0\,,$$
and the equality holds iff $a=0$.  You can also use the derivative form:
$$(1+a)^n-\sum_{r=0}^3\,\binom{n}{r}\,a^r=\binom{n}{k+1}\,(1+\zeta)^{n-k-1}\,a^4\geq0\,,$$
where $\zeta$ is a number between $0$ and $a$.
A: proof-verification:

Proof. From the binomial theorem we can see that, 
  $$
(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,
$$ 
  becomes ("is equivalent to"), 
  $$
\sum^{n}_{k=0}\binom{n}{k}a^{k}\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}.
$$ 
  Also, if $a\lt b$ (I would use other notations here so that I would not mess up with the "a" in the inequality), let us define $\sum^{a}_{i=b}f(i)=0$, by convention.
I would not bother with the cases n=1,2,3: one can just claim that one can just do it by direct substitution.
More generally Suppose $n\geqslant 4$. Then it suffices to show that (you don't need to write so much here. I have tailored some of your steps.) 
  $$
\sum^{n}_{k=4}\binom{n}{k}a^{k}\geq 0
$$
  which is
  $$
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}\ge0.
$$

To determine whether this last statement is true, We first consider $a\ge0$, and we see that it is trivially true. Next we consider the case where $-1\le a\lt0.$ We let $a=-b$ (I would write b=-a instead. When defining something new, it is clearer to write it on the left.), so $0\lt b\le 1$ and notice that, 
  $$
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}
\\
=\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^{k}\binom{n}{k}b^{k}+...+(-1)^{n}\binom{n}{n}b^{n}.
$$
Also from the binomial theorem we can see that (A possible big gap here: this step is mysterious to me and totally unmotivated: how do you suddenly get rid of the a and b in the equality above and jump to the following one? I would stop reading here.), 
  \begin{align}
(1+(-1))^{n}-(1+(-1))^{3}+(1+(-1))^{1}\\
=\sum^{n}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}-\sum^{3}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}+\sum^{1}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}\\ 
0=1-\binom{n}{1}+\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\\ 
n-1=\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\ge0,
\end{align} 
  and by noticing that $b^4\ge b^5\ge b^6\ge ...\ge b^k$, we can see that each binomial term $\binom{n}{k}$ is multiplied by a factor of $b^k$, making each term smaller than the term before. 
Thus (Assuming the above is correct, I don't understand this step either.),
  \begin{align}
\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^n\binom{n}{n}b^n\ge0\\ 
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{n}a^n\ge0,\tag{by substitution}
\end{align} as desired.
Working backward(the rest is redundant. Writing could be precise without sacrificing any rigorousness.)

A: Also, we can use the following reasoning.
For $n\in\{1,2,3\}$ it's an identity.
Let $n\geq4$ and $$f(a)= (1+a)^n-1-na-\frac{n(n-1)}{2}a^{2}-\frac{n(n-1)(n-2)}{6}a^3.$$
Thus, 
$$f'(a)=n(1+a)^n-n-n(n-1)(n-2)a-\frac{n(n-1)(n-2)}{2}a^2,$$,
$$f''(a)=n(n-1)(1+a)^{n-2}-n(n-1)-n(n-1)(n-2)a$$ and
$$f'''(a)=n(n-1)(n-2)(1+a)^{n-3}-n(n-1)(n-2).$$
We see that $f''$ gets a minimal value for $a=0$.
Thus, $$f''(a)\geq f(0)=0,$$
which gives that $f'($ is an increasing function and $f'(0)=0$.
Thus, since $f'$ has an unique root on $[-1,+\infty)$, we see that $x_{min}=0$,
which gives $f(x)\geq f(0)=0$ and we are done! 
