Show that $$1^3+2^3+3^3+\text{...}+n^3=(1+2+3+\text{...}+n)^2$$ For $n=1$ we have $$1^3=1^2$$ which is obviously true. Assume that $$1^3+2^3+3^3+\text{...}+k^3=(1+2+3+\text{...}+k)^2$$is true for some positive integer $k\ge1$. We shall now prove that it is also true for $n=k+1$. I don't see how can we do that. $$1^3+2^3+3^3+\text{...}+k^3+(k+1)^3\overset{?}{=}(1+2+3+\text{...}+k+(k+1))^2$$ The LHS is $$(1+2+3+\text{...}+k)^2+(k+1)^3=...$$
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Hint: it's a lot easier if you can assume (or show by induction) that $1 + 2 + \dots + k = \frac 12 k(k+1)$ first. – Deepak Jan 16 '22 at 11:39
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A more interesting question, IMHO is: find a geometric proof of this statement. – Igor Rivin Jan 16 '22 at 11:43
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See for example https://math.stackexchange.com/q/1125766/42969 or https://math.stackexchange.com/q/1111443/42969 or https://math.stackexchange.com/q/996083/42969 – Martin R Jan 16 '22 at 11:45
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1For a visual proof, see https://math.stackexchange.com/q/3938027/42969. – Martin R Jan 16 '22 at 11:46
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1For what it's worth, I posted an answer and then deleted it, because my answer is a duplicate. – user2661923 Jan 16 '22 at 11:57
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@EliasCosta, obviously I wasn't familiar with the fact that $1+2+\text{...}+k=\dfrac12k(k+1)$, so no. – Math Student Jan 16 '22 at 12:02
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You need to show that $1^3 + 2^3 + \dots + k^3 + (k+1)^3 = (1+2+ \dots + k+k+1)^2$.
It's a lot easier if you can assume (or show by induction) that $1 + 2 + \dots + k = \frac 12 k(k+1)$ first.
Using that result,
$1^3 + 2^3 + \dots + k^3 + (k+1)^3 \\= (\frac 12k(k+1))^2 + (k+1)^3 \\= (\frac12)^2(k+1)^2(k^2 + 4k +4) \\= (\frac12)^2(k+1)^2(k+2)^2 \\= (1 + 2 + \dots + k+k+1)^2$
as required.
Deepak
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Thank you for the response! Can you clarify how $$(\frac12)^2(k+1)^2(k+2)^2)$$ is equal to the RHS? – Math Student Jan 16 '22 at 12:01
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See the duplicate for an induction proof. This question has been asked several times already. – Dietrich Burde Jan 16 '22 at 12:03
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1@Medi That's equal to $(\frac 12(k+1)(k+2))^2 = (1 + 2 + \dots + k+1)^2$. It's basically reversing the lemma I asked you to show first. Do you see? – Deepak Jan 16 '22 at 12:10
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@DietrichBurde Sorry, I get your point, and what you wrote is true. I did want to help the OP out as they wrote a good quality question and gave good details of their doubt. – Deepak Jan 16 '22 at 12:11
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@DietrichBurde, if you can't help me, I really don't see what's the point of you writing useless comments under people's tries to assist me. – Math Student Jan 16 '22 at 12:15
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Here's a beautiful visual proof. I got the picture from Google.
The blocks on the left are of course cubes and not squares as they may look like.
Sayan Dutta
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@DietrichBurde no, I didn't see that. That one's nicer of course. But, that's no reason for the downvote- my answer is correct and I put my own effort in it. – Sayan Dutta Jan 16 '22 at 17:41