$$ \sum_{k=1}^n k^3 = \left( \sum_{k=1}^n k \right)^2 $$ I can't quite understand this expression, and in fact this is my biggest difficulty in finding a solution. Can someone please explain to me ? $$ \sum_{k=1}^n k = \frac{n(n+1)} 2 $$
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2Left side is $1^3+2^3+\cdots+n^3$. Right side is $(1+2+\cdots+n)^2$ – paw88789 Nov 06 '14 at 18:47
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I love this result: it's quite simple and looks like it should be wrong. – Simon S Nov 06 '14 at 18:52
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See also: http://math.stackexchange.com/questions/973456/prove-by-induction-sum-n3-sum-n2 – apnorton Nov 06 '14 at 19:13
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It's asking you to prove that the following identity holds for all $n \in \mathbb N$: $$ 1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2 $$ For example, when $n = 5$, we have that: $$ 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 = (1 + 2 + 3 + 4 + 5)^2 $$
Adriano
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1@MatematikaMatematika: have you used induction previously? What would be the base step? How would the assumption step look? – abiessu Nov 06 '14 at 18:50
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i guess i prove for n=1 then we assume for n=k it's true and then by using this assumption we prove it for n=k+1 – Matematika Matematika Nov 06 '14 at 18:51
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1@MatematikaMatematika Have you already seen a closed form expression for $1 + 2 + \cdots + n$? That can help. – Adriano Nov 06 '14 at 18:52
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you must prove that $1^3+2^3+...+n^3+(n+1)^3=(1+2+...+n+(n+1))^2$ and use that $1^3+2^3+...+n^3=(1+2+...+n)^2$ holds. Now you can write $1^3+2^3+3^3+...+n^3+(n+1)^3=(1+2+...+n)^2+(n+1)^3$ in the next step we have to show that $(1+2+...+n)^2+(n+1)^3=(1+2+...+n+(n+1))^2$
Dr. Sonnhard Graubner
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You have $S_n=1+...+n = {1 \over 2} n (n+1)$. This is straightforward to show.
Let $C_n = 1^3+... +n^3$.
You want to show $C_n = S_n^2$. It is true for $n=1$, so suppose it is true for $n$.
Then $C_{n+1} = C_n + n^3$.
Also, $S_{n+1}^2 = (S_n + (n+1))^2 = S_n^2 +2 S_n (n+1) + (n+1)^2$.
Note that $2 S_n (n+1) + (n+1)^2 = (n+1)^2 (n+1) = (n+1)^3 $.
Hence $ C_{n+1} = S_{n+1}^2$.
copper.hat
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