Help in easy floor summation

How do I calculate the following summation? \[ \sum_{i = 1}^{\lfloor \sqrt{n} \rfloor} \Bigl \lfloor \frac{n}{i^2} \Bigr \rfloor i^2 \] \[ \sum_{i = 1}^{\lfloor \sqrt[3]{n} \rfloor} \Bigl \lfloor \frac{n}{i^3} \Bigr \rfloor i^3 \]

and generalize for any power?

#NumberTheory

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

$\left\lfloor\frac{N}{i^2}\right\rfloor=1$ whenever $1\leq\frac{N}{i^2}<2$ The sum can be reduced like this even further. $S= \sum_{1}^{\bigl \lfloor \sqrt{\frac{n}{2}} \bigr \rfloor} \Bigl \lfloor \frac{n}{i^2} \Bigr \rfloor i^2 + \sum_{i = \bigl \lfloor \sqrt{\frac{n}{2}} \bigr \rfloor}^{\lfloor \sqrt{n} \rfloor} 1 . i^2$ @Calvin Lin @Agnishom Chattopadhyay Maybe you guys can help further.

Rishabh Deep Singh - 1 year, 2 months ago

This is from a live Code Chef problem.

Calvin Lin Staff - 1 year, 2 months ago

it is over now we can discuss it.

Rishabh Deep Singh - 1 year, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$