I have the following polynomial defined : $P_k = (x+1)^{k+1} - x^{k+1}$ with $k \in {0,1,2,...,n}$
I need to prove that the set of polynomials : $(P_0,P_1,...,P_n)$ is a basis for $\mathbb{K_n}[X]$
My Proof:
I have that $$dim \mathbb{K_n}[X] = card(P_0,P_1,P_2,...,P_n) = n+1$$ so I just need to prove that the polynomials $P_i$ are linearly independent, so I proved that $$\sum_{0}^{n} a_iP_i = 0 \rightarrow a_i = 0$$
To do this, I used a basic induction :
for $i = 0$ :
$a_0P_0 = a_0(x+1) - a_0x$,it's trivial that in this case $a_0 = 0$, then I supposed that for every natural number $n$ we have the following : $\sum_{0}^{n} a_iP_i = 0 \rightarrow a_i = 0$ and I have to prove it for $n+1$ :
$$ \sum_{0}^{n+1} a_iP_i = 0 \rightarrow \sum_{0}^{n} a_iP_i + a_{n+1}P_{n+1} = 0 $$
Since : $\sum_{0}^{n} a_iP_i = 0 \rightarrow a_i = 0$ then we easily get :
$$a_{n+1}P_{n+1} = a_{n+1}(x + 1)^{n+2} - a_{n+1}x^{n+2} = 0$$ , I couldn't find a good way to prove that this leads to $a_{n+1} = 0$ execpt from saying that the equality : $a_{n+1}(x + 1)^{n+2} = a_{n+1}x^{n+2}$ doesn't hold except if $a_{n+1} = 0$ , but somehow I don't find this workaround elegant, Is this induction correct ? if so how to finish it ? Is there any way to prove that it's a basis without induction ?
What you have to prove is that$$\sum_{i=0}^{n+1}a_iP_i=0\implies(\forall i\in\{0,1,\ldots,n+1\}):a_i=0.$$But ithe coefficient of $x^{n+1}$ in the polynomial $\sum_{i=0}^{n+1}a_iP_i$ is $a_{n+1}$. So, if this polynial is the null poynomial, then its coefficients are $0$ and in particular, $a_{n+1}=0$. So, since $\sum_{i=0}^{n+1}a_iP_i=0$ and since $a_{n+1}=0$, now you know that $\sum_{i=0}^na_iP_i=0$ and therefore you can apply the induction hypothesis.