[proofplan] We prove the partition-of-unity identity by induction on $n$, using only the Pascal recurrence for binomial coefficients and the fact that $x+(1-x)=1$. The induction step rewrites the degree-$(n+1)$ Bernstein sum as $(1-x)S_n(x)+xS_n(x)$, where $S_n$ is the degree-$n$ Bernstein sum. Non-negativity is then immediate from the definition of each Bernstein polynomial, since every factor in the product is non-negative on $[0,1]$. [/proofplan]

[step:Define the Bernstein sum and prove the base case] For each $n \in \mathbb{N}$, define the Bernstein sum \begin{align*} S_n: [0,1] \to \mathbb{R} \end{align*} by \begin{align*} S_n(x) = \sum_{k=0}^{n} \binom{n}{k}x^k(1-x)^{n-k}. \end{align*} For $n=1$ and $x \in [0,1]$, we compute directly: \begin{align*} S_1(x) = \binom{1}{0}(1-x) + \binom{1}{1}x = 1-x+x = 1. \end{align*} Thus the desired identity holds for $n=1$. [/step]

[step:Rewrite the degree-$(n+1)$ sum using the degree-$n$ sum]Assume that for some $n \in \mathbb{N}$ one has $S_n(x)=1$ for every $x \in [0,1]$. Fix $x \in [0,1]$. By the Pascal identity \begin{align*} \binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1} \end{align*} for $1 \leq k \leq n$, together with the boundary identities $\binom{n+1}{0}=\binom{n}{0}$ and $\binom{n+1}{n+1}=\binom{n}{n}$, the degree-$(n+1)$ Bernstein sum decomposes as \begin{align*} S_{n+1}(x)=\sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n+1-k}+\sum_{k=0}^{n}\binom{n}{k}x^{k+1}(1-x)^{n-k}. \end{align*} Factoring out the common powers gives \begin{align*} S_{n+1}(x)=(1-x)\sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n-k}+x\sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n-k}. \end{align*} Therefore \begin{align*} S_{n+1}(x)=(1-x)S_n(x)+xS_n(x). \end{align*}[/step]

[guided]The purpose of this step is to express the degree-$(n+1)$ sum in terms of the degree-$n$ sum, so that the induction hypothesis can be used. Fix $x \in [0,1]$ and define \begin{align*} S_n(x)=\sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n-k}. \end{align*} We start from the degree-$(n+1)$ expression: \begin{align*} S_{n+1}(x)=\sum_{k=0}^{n+1}\binom{n+1}{k}x^k(1-x)^{n+1-k}. \end{align*} For the interior indices $1 \leq k \leq n$, Pascal's identity gives \begin{align*} \binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}. \end{align*} At the endpoints, the same decomposition is represented by the boundary terms $\binom{n+1}{0}=\binom{n}{0}$ and $\binom{n+1}{n+1}=\binom{n}{n}$. Hence the whole sum splits into two sums: \begin{align*} S_{n+1}(x)=\sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n+1-k}+\sum_{k=0}^{n}\binom{n}{k}x^{k+1}(1-x)^{n-k}. \end{align*} The first sum has one extra factor of $1-x$ compared with $S_n(x)$: \begin{align*} \sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n+1-k}=(1-x)S_n(x). \end{align*} The second sum has one extra factor of $x$ compared with $S_n(x)$: \begin{align*} \sum_{k=0}^{n}\binom{n}{k}x^{k+1}(1-x)^{n-k}=xS_n(x). \end{align*} Combining these two identities gives \begin{align*} S_{n+1}(x)=(1-x)S_n(x)+xS_n(x). \end{align*} This is exactly the form needed for induction, because the right-hand side contains only $S_n(x)$.[/guided]

[step:Apply the induction hypothesis to obtain the partition identity] By the induction hypothesis, $S_n(x)=1$. Therefore \begin{align*} S_{n+1}(x)=(1-x)S_n(x)+xS_n(x)=(1-x)\cdot 1+x\cdot 1=1. \end{align*} Since $x \in [0,1]$ was arbitrary, $S_{n+1}(x)=1$ for every $x \in [0,1]$. By induction, for every $n \in \mathbb{N}$ and every $x \in [0,1]$, \begin{align*} \sum_{k=0}^{n}\binom{n}{k}x^k(1-x)^{n-k}=1. \end{align*} [/step]

[step:Check non-negativity of each Bernstein basis polynomial] Fix $n \in \mathbb{N}$, $k \in \{0,\dots,n\}$, and $x \in [0,1]$. Since $x \geq 0$ and $1-x \geq 0$, the powers $x^k$ and $(1-x)^{n-k}$ are non-negative. Also $\binom{n}{k} \geq 0$. Hence \begin{align*} B_{n,k}(x)=\binom{n}{k}x^k(1-x)^{n-k}\geq 0. \end{align*} Together with the partition identity proved above, this proves that the Bernstein basis polynomials form a non-negative [partition of unity](/page/Partition%20of%20Unity) on $[0,1]$. [/step]