MHB Can you prove the Floor Function Challenge?

AI Thread Summary
The discussion centers on proving the equation involving the floor function: $$\sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor=\left\lfloor{\dfrac{n(n+2)(2n-1)}{24}}\right\rfloor. Participants share methods for proving the statement, starting with the case when n is even, and then extending the proof to odd n by considering the last n-1 terms and the nth term. Acknowledgment is given to kaliprasad for their solution, indicating a collaborative effort in problem-solving. The discussion highlights the mathematical reasoning behind the proofs and the shared approaches among contributors. Overall, the thread emphasizes the importance of both even and odd cases in validating the floor function challenge.
anemone
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Prove $$\sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor=\left\lfloor{\dfrac{n(n+2)(2n-1)}{24}}\right\rfloor$$.
 
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anemone said:
Prove $$\sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor=\left\lfloor{\dfrac{n(n+2)(2n-1)}{24}}\right\rfloor$$.
I shall prove the same for n even
we have for k even

$\lfloor (\dfrac{k}{2})^2 \rfloor = (\dfrac{k}{2})^2\cdots(1)$ because $ (\dfrac{k}{2})$ is integer

further we have for k odd say (2m-1)

$\lfloor (\dfrac{k}{2})^2 \rfloor$

= $\lfloor (\dfrac{2m-1}{2})^2 \rfloor$

= $\lfloor \dfrac{4m^2-4m +1}{4} \rfloor$

= $\lfloor m^2-m + \dfrac{1}{4} \rfloor$

= $m^2-m$so summing the given expression from k = 1 to n breaking into even part and odd part we get

$\sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor$

= $\sum_{k=1}^{\frac{n}{2}}\left\lfloor{\left(\frac{2k-1}{2}\right)^2+ \left(\frac{2k}{2}\right)^2}\right\rfloor$ justified to merge the 2 because 2nd expression is integer

= $\sum_{k=1}^{\frac{n}{2}}{k^2-k+ k^2}$

= $\sum_{k=1}^{\frac{n}{2}}{2k^2-k}$

= $2 * \dfrac{\frac{n}{2}\cdot(\frac{n}{2}+1)\cdot(n+1)}{6}- \dfrac{\frac{n}{2} \cdot (\frac{n}{2}+1)}{2}$

= $\dfrac{n\cdot(n+2)\cdot(n+1)}{12}- \dfrac{n\cdot(n + 2)}{8}$

= $n\cdot(n+2)(\dfrac{2n-1}{24})$

= $\dfrac{n\cdot(n+2)\cdot(2n-1)}{24}$



becuase above is integer ( being sum of integers) hence same as $\lfloor \dfrac{n\cdot(n+2)\cdot(2n-1)}{24}\rfloor $

hence the result
 
kaliprasad said:
I shall prove the same for n even
we have for k even

$\lfloor (\dfrac{k}{2})^2 \rfloor = (\dfrac{k}{2})^2\cdots(1)$ because $ (\dfrac{k}{2})$ is integer

further we have for k odd say (2m-1)

$\lfloor (\dfrac{k}{2})^2 \rfloor$

= $\lfloor (\dfrac{2m-1}{2})^2 \rfloor$

= $\lfloor \dfrac{4m^2-4m +1}{4} \rfloor$

= $\lfloor m^2-m + \dfrac{1}{4} \rfloor$

= $m^2-m$so summing the given expression from k = 1 to n breaking into even part and odd part we get

$\sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor$

= $\sum_{k=1}^{\frac{n}{2}}\left\lfloor{\left(\frac{2k-1}{2}\right)^2+ \left(\frac{2k}{2}\right)^2}\right\rfloor$ justified to merge the 2 because 2nd expression is integer

= $\sum_{k=1}^{\frac{n}{2}}{k^2-k+ k^2}$

= $\sum_{k=1}^{\frac{n}{2}}{2k^2-k}$

= $2 * \dfrac{\frac{n}{2}\cdot(\frac{n}{2}+1)\cdot(n+1)}{6}- \dfrac{\frac{n}{2} \cdot (\frac{n}{2}+1)}{2}$

= $\dfrac{n\cdot(n+2)\cdot(n+1)}{12}- \dfrac{n\cdot(n + 2)}{8}$

= $n\cdot(n+2)(\dfrac{2n-1}{24})$

= $\dfrac{n\cdot(n+2)\cdot(2n-1)}{24}$



becuase above is integer ( being sum of integers) hence same as $\lfloor \dfrac{n\cdot(n+2)\cdot(2n-1)}{24}\rfloor $

hence the result
now based on result of even we prove for odd ( last n-1 terms + nth term , n is odd)
we have shown that for k odd = 2m- 1
$\lfloor (\dfrac{k}{2})^2 \rfloor$
= $\lfloor (\dfrac{2m-1}{2})^2 \rfloor$
= $\lfloor \dfrac{4m^2-4m +1}{4} \rfloor$
= $\lfloor m^2-m + \dfrac{1}{4} \rfloor$
= $m^2-m$
= $m(m-1)$
= $\dfrac{n+1}{2} \dfrac{n-1}{2}$
= $\dfrac{n^2-1}{4}$Now sum upto n terms

= $sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor$
= $(sum_{k=1}^{n-1}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor)+ \left\lfloor{\left(\frac{n}{2}\right)^2}\right\rfloor$
= $\dfrac{(n-1)(n+1)(2n-3)}{24} + \dfrac{n^2-1}{4}$ (sum upto n-1 terms n-1 even taken care)
= $\dfrac{(n-1)(n+1)(2n-3)+6n^2-6}{24}$
= $\dfrac{(n^2-1)(2n-3)+6n^2-6}{24}$
= $\dfrac{2n^3-3n^2-2n + 3+6n^2-6}{24}$
= $\dfrac{2n^3+ 3n^2-2n - 3}{24}$
= $\dfrac{n(2n^2 + 3n-2) - 3}{24}$
= $\dfrac{n(n+2)(2n-1)}{24} -\dfrac{1}{8}$
= $\lfloor \dfrac{n(n+2)(2n-1)}{24} \rfloor$
 
Thank you kaliprasad for your solution, very well done! My approach is more or less the same as yous, just so you know. :)
 
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