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anemone
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Prove \(\displaystyle \sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor=\left\lfloor{\dfrac{n(n+2)(2n-1)}{24}}\right\rfloor\).
I shall prove the same for n evenanemone said:Prove \(\displaystyle \sum_{k=1}^{n}\left\lfloor{\left(\frac{k}{2}\right)^2}\right\rfloor=\left\lfloor{\dfrac{n(n+2)(2n-1)}{24}}\right\rfloor\).
now based on result of even we prove for odd ( last n-1 terms + nth term , n is odd)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
The Floor Function Challenge is a mathematical problem that involves finding the largest integer less than or equal to a given real number. It is often used in computer programming and can be solved using the floor function, which rounds down a number to the nearest integer.
The floor function is represented using the notation "⌊x⌋" or "floor(x)", where x is the number being rounded down. It can also be written as "int(x)" or "integer(x)".
Yes, the floor function can be used with both positive and negative numbers. When applied to a negative number, the floor function will round the number down to the nearest integer, which is closer to negative infinity.
The floor function rounds numbers down to the nearest integer, while the ceiling function rounds numbers up to the nearest integer. For example, the floor of 3.5 would be 3, while the ceiling of 3.5 would be 4.
The Floor Function Challenge is commonly used in computer programming to solve problems that involve rounding down to the nearest integer. It can also be used in various mathematical and scientific calculations where precise integer values are required.