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I tried using induction:

Let $x =[\sqrt 1, \sqrt 2]$. Then by Pythagoras $\|x\| = \sqrt 3$. Suppose $\|u\|$ is defined for $u$ in $\mathbb R^{\sqrt n}$. Then $\|u\| = \sqrt{1 + 2 + 3 + … + n}$. Since the length is defined for $\mathbb R^2$, $\|y\| = \sqrt{(1 + 2 + 3 + … + n) + (n + 1)}$ is well-defined. But that's the length of $u$.

Please, check and see what I need to do to fix/improve it. Thanks.

edit:

$u \in \mathbb R^n.$

Length in $\mathbb R^2$ is given by Pythagoras.

$\|u\|$ is well-defined by definition of length.

numbers
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    $u$ in $\mathbb R^{\sqrt n}$? – The length/dimension of the vector $u$ is $n$, not $\sqrt n$. – Martin R Dec 19 '15 at 18:09
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    It seems like you have the main idea (i.e. the expression for $|u|$), but a lot of what you've written doesn't make sense. What is $\Bbb R^{\sqrt n}$? What do you mean by "suppose $|u|$ is defined"? What do you mean by "since length is defined for $\Bbb R^2$?" What do you mean by $|y| = \cdots$ "is well-defined"? It reads like a mathematical word salad. – Ben Grossmann Dec 19 '15 at 18:11
  • I edited my OP. – numbers Dec 19 '15 at 18:18
  • See also: http://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-ldotsn – Martin Sleziak Dec 20 '15 at 08:18

1 Answers1

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\begin{align} \lVert u \rVert^2 = \lVert u \rVert_2^2 &= \sum_{i=1}^n u_i^2 \\ &= \sum_{i=1}^n (\sqrt{i})^2 \\ &= \sum_{i=1}^n i \\ &\overset{(*)}{=} \frac{n(n+1)}{2} \end{align} So we get $$ \lVert u \rVert = \sqrt{\frac{n(n+1)}{2}} $$

mvw
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