[proofplan] We first prove the two-vector case by expanding $\|v + w\|^2$ using the bilinearity of the inner product, observing that orthogonality kills the cross terms. The general case follows by induction on $k$, applying the two-vector case at each inductive step after verifying that the partial sum remains orthogonal to the next vector. [/proofplan]

[step:Expand $\|v + w\|^2$ using bilinearity and apply orthogonality]By definition of the induced norm, $\|v + w\|^2 = \langle v + w, v + w \rangle$. Expanding by bilinearity of the inner product in both arguments: \begin{align*} \|v + w\|^2 &= \langle v + w, v + w \rangle \\ &= \langle v, v \rangle + \langle v, w \rangle + \langle w, v \rangle + \langle w, w \rangle. \end{align*} Since $v \perp w$, we have $\langle v, w \rangle = 0$. By symmetry of the inner product on a real vector space, $\langle w, v \rangle = \langle v, w \rangle = 0$. Therefore \begin{align*} \|v + w\|^2 &= \langle v, v \rangle + \langle w, w \rangle = \|v\|^2 + \|w\|^2. \end{align*}[/step]

[guided]We want to show $\|v + w\|^2 = \|v\|^2 + \|w\|^2$. The norm $\|x\|$ on an inner product space is defined by $\|x\|^2 = \langle x, x \rangle$, so the natural approach is to expand the left-hand side using the inner product. By definition, $\|v + w\|^2 = \langle v + w, v + w \rangle$. The inner product on a real vector space is bilinear (linear in each argument) and symmetric ($\langle x, y \rangle = \langle y, x \rangle$). Expanding: \begin{align*} \|v + w\|^2 &= \langle v + w, v + w \rangle \\ &= \langle v, v \rangle + \langle v, w \rangle + \langle w, v \rangle + \langle w, w \rangle. \end{align*} Now we use the hypothesis $v \perp w$, which means $\langle v, w \rangle = 0$. Since the inner product is symmetric, $\langle w, v \rangle = \langle v, w \rangle = 0$ as well. The two cross terms vanish, leaving: \begin{align*} \|v + w\|^2 &= \langle v, v \rangle + \langle w, w \rangle = \|v\|^2 + \|w\|^2. \end{align*} This completes the two-vector case. The key point is that orthogonality eliminates the cross terms $\langle v, w \rangle$ and $\langle w, v \rangle$, which are the only terms preventing the norm from being additive.[/guided]

[step:Extend to $k$ pairwise orthogonal vectors by induction]We proceed by induction on $k$. **Base case ($k = 1$).** The identity $\|v_1\|^2 = \|v_1\|^2$ holds. **Inductive step.** Assume the identity holds for some $k - 1 \geq 1$ pairwise orthogonal vectors. Let $v_1, \ldots, v_k \in V$ be pairwise orthogonal. Define $s_{k-1} := \sum_{i=1}^{k-1} v_i$. By the inductive hypothesis, \begin{align*} \|s_{k-1}\|^2 &= \sum_{i=1}^{k-1} \|v_i\|^2. \end{align*} We verify that $s_{k-1} \perp v_k$. By linearity of the inner product in the first argument: \begin{align*} \langle s_{k-1}, v_k \rangle &= \left\langle \sum_{i=1}^{k-1} v_i,\, v_k \right\rangle = \sum_{i=1}^{k-1} \langle v_i, v_k \rangle = 0, \end{align*} where the last equality holds because $v_i \perp v_k$ for each $i \in \{1, \ldots, k-1\}$ by the pairwise orthogonality hypothesis. Applying the two-vector result established above to $s_{k-1}$ and $v_k$: \begin{align*} \left\|\sum_{i=1}^k v_i\right\|^2 = \|s_{k-1} + v_k\|^2 = \|s_{k-1}\|^2 + \|v_k\|^2 = \sum_{i=1}^{k-1} \|v_i\|^2 + \|v_k\|^2 = \sum_{i=1}^k \|v_i\|^2. \end{align*} This completes the inductive step and the proof.[/step]

[guided]To extend from two vectors to $k$ pairwise orthogonal vectors, we use induction on $k$. The idea is to group the first $k - 1$ vectors into a single partial sum and then apply the two-vector result to this partial sum and the $k$-th vector. **Base case ($k = 1$).** The statement $\|v_1\|^2 = \|v_1\|^2$ holds. **Inductive step.** Suppose the identity holds for any collection of $k - 1$ pairwise orthogonal vectors. Let $v_1, \ldots, v_k \in V$ be pairwise orthogonal. Define the partial sum $s_{k-1} := \sum_{i=1}^{k-1} v_i$. Since $v_1, \ldots, v_{k-1}$ are pairwise orthogonal (a subset of the original collection inherits pairwise orthogonality), the inductive hypothesis gives \begin{align*} \|s_{k-1}\|^2 &= \sum_{i=1}^{k-1} \|v_i\|^2. \end{align*} To apply the two-vector case to $s_{k-1}$ and $v_k$, we must verify that $s_{k-1} \perp v_k$. Using linearity of the inner product in the first argument: \begin{align*} \langle s_{k-1}, v_k \rangle &= \left\langle \sum_{i=1}^{k-1} v_i,\, v_k \right\rangle = \sum_{i=1}^{k-1} \langle v_i, v_k \rangle. \end{align*} Each term $\langle v_i, v_k \rangle = 0$ because $v_i \perp v_k$ for $i \in \{1, \ldots, k-1\}$ by the pairwise orthogonality hypothesis. So $\langle s_{k-1}, v_k \rangle = 0$, confirming $s_{k-1} \perp v_k$. Now the two-vector Pythagorean identity applies: \begin{align*} \left\|\sum_{i=1}^k v_i\right\|^2 &= \|s_{k-1} + v_k\|^2 = \|s_{k-1}\|^2 + \|v_k\|^2 = \sum_{i=1}^{k-1} \|v_i\|^2 + \|v_k\|^2 = \sum_{i=1}^k \|v_i\|^2. \end{align*} This completes the induction. The critical verification is that the partial sum $s_{k-1}$ is orthogonal to $v_k$ -- this is where pairwise orthogonality (not just sequential orthogonality) is used, since we need $\langle v_i, v_k \rangle = 0$ for every $i < k$.[/guided]