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This problem was given in a math contest, and the following answers were provided

  • $\sum_{k=1}^n4k^3$
  • $\sum_{k=1}^n2k^3$
  • $\sum_{k=1}^n6k^2$
  • $\sum_{k=1}^n4k^2$
  • $\sum_{k=1}^n2k^4$
  • none of the above

Thanks!

EDIT: I'm sorry, the first term should be n^4.

OK so what I've considered so far is factoring down the expression to $(n^2)(n+1)^2$

From there, I'm not sure if there's a way to recognize that that's a summation or something (Similar to $n*(n+1)/2$)

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    This site works a lot better if you share your own thoughts on the problem, first. – 2017-01-28
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    I'm sorry; I wish I could, but honestly I barely understand this. I would appreciate if someone could provide an easier example or something. – 2017-01-28

5 Answers 5

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Note that :$$\color{red}{1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2=(\dfrac{n(n+1)}{2})^2}$$$$n^4+2n^3+n^2=n^2(n^2+2n+1)=n^2(n+1)^2=\dfrac44n^2(n+1)^2=\\4((\dfrac{n(n+1)}{2})^2=\\4(1^3+2^3+3^3+4^3+...+n^3)=\\4\sum_{k=1}^{n}k^3$$ and...

this is mackett enter image description here

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    That's a very nice image, but strictly speaking, you're missing a proof there. – 2017-01-28
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    This my handmade proof verification .I usually use it in my class when I'm teaching . – 2017-01-29
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Plugging in $n=1$ and $n=2$ shows that only $\sum_{k=1}^n4k^3$ is a candidate. So we need to show that $\sum_{k=1}^n4k^3 = n^4+2n^3+n^2$

Proof by induction:

$n=1$: already done

$n\to n+1$:

$$\sum_{k=1}^{n+1}4k^3 - ((n+1)^4+2(n+1)^3+(n+1)^2)\\= \sum_{k=1}^{n}4k^3+4(n+1)^3 - (n^4+4n^3+6n^2+4n+1+2(n^3+3n^2+3n+1)+n^2+2n+1)\\= n^4+2n^3+n^2+4(n^3+3n^2+3n+1) - (n^4+6n^3+13n^2+12n+4)\\=0$$

Note that with this kind of series it is easier to show $l-r=0$ instead of $l=…=r$.

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You are looking for a function $g(k)$ (should it exist) such that

$$f(n)=\sum_{k=1}^ng(k)=[n(n+1)]^2$$

Note that $f(n)-f(n-1)=g(n)$. So we have

$$f(n)-f(n-1)=(n^2+n)^2-(n^2-n)^2$$

The only terms that don't cancel out are the $n^3$ terms, leaving $g(n)=4n^3$. Note that this doesn't actually prove $g(k)$ exists, just that it can't be anything else.

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Answer: \begin{equation} \sum_{i=1}^n{4k^3} \end{equation}

Short Explanation:

you can identify that with the series $\sum_{i=1}^n{k^3} = n^4/4 + n^3/2 + n^2/4$

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There are many solutions. For example, using the binomial theorem:

$$n^4+2n^3+n^2=n^2(1+n)^2=n^2\sum_{k=0}^2\binom2kn^k=\boxed{\sum_{k=0}^2\binom2kn^{k+2}}$$