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I am trying to prove by induction that: $$1^3-2^3 +\cdots+n^3=(1+2+\cdots+n)^2 $$

This was a problem from a practice worksheet, but I don't understand how to interpret the LHS.

Is the following correct:

$$1^3-2^3+3^3-4^3+5^3-6^3+\cdots+n^3$$

Or this:

$$1^3-2^3+3^3+4^3+5^3+6^3+\cdots+n^3$$

I have a feeling it is the former. If this is the case, I presumably have to consider odd and even cases for n right?

Thanks

Arturo Magidin
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subscript42
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    It's a typo; they should all be plus signs. Otherwise, it's not true. – Arturo Magidin Dec 22 '21 at 20:29
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    It is true that using $\dots$ can easily lead to ambiguity. Yet people keep using them because they are sometimes more intuitive than any other ways of expression. In this case however, it's simply a typo: the $-$ should be $+$ for the equality to hold. – WhatsUp Dec 22 '21 at 20:30
  • Ahh that makes sense! Thank you very much – subscript42 Dec 22 '21 at 20:31
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    Aside from the truth of the statement, the first one looks far more attractive. – Hermis14 Dec 22 '21 at 20:31
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    If were really to be an alternating sign, the least term would have had to be written something like $(-1)^{n+1}n^3$ or something like that. – Arturo Magidin Dec 22 '21 at 20:34
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    You should approach problems like this by calculating some small cases to see whether this gives you an idea. It may also catch typos. In the case n=2 the top line LHS is negative and the RHS is positive. – DanielWainfleet Dec 23 '21 at 04:58

1 Answers1

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As pointed out, the true formula is that

$1^3+2^3+...+n^3=(1+2+...+n)^2$,

and you can prove this by induction. Assume that

$1^3+2^3+\cdots +(n-1)^3=(1+2+\cdots +(n-1))^2$,

then

$1^3+2^3+\cdots +n^3=(1+2+...+n-1)^2+n^3=\frac{n^2(n-1)^2}{4}+n^3=(1+2+\cdots n)^2.$

simonc
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