Can you prove the Floor Function Challenge?

[anemone]

Prove $$\sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor=\left\lfloor{\dfrac{n(n+2)(2n-1)}{24}}\right\rfloor$$.

 

[kaliprasad]

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

Spoiler

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

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

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

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

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

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

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



because 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]

I shall prove the same for n even

Spoiler

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

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

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

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

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

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

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



because 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)

Spoiler

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)
= $\dfrac{(n-1)(n+1)(2n-3)+6n^2-6}{24}$
= $\dfrac{(n^2-1)(2n-3)+6n^2-6}{24}$
= $\dfrac{2n^3-3n^2-2n + 3+6n^2-6}{24}$
= $\dfrac{2n^3+ 3n^2-2n - 3}{24}$
= $\dfrac{n(2n^2 + 3n-2) - 3}{24}$
= $\dfrac{n(n+2)(2n-1)}{24} -\dfrac{1}{8}$
= $\lfloor \dfrac{n(n+2)(2n-1)}{24} \rfloor$

 

[anemone]

Thank you kaliprasad for your solution, very well done! My approach is more or less the same as yous, just so you know. :)