# Pies all over the floor!

Hello everybody! This is my very first note. As the title suggests, today we are going to talk about something related to pi. Nah, not the chicken pie or apple pie or any other pies that you enjoy eating but the pi defined as the ratio of circumference to the diameter of a circle!

Well, to get things started, find $\lfloor \pi\rfloor+\lfloor \pi^{2}\rfloor+\lfloor \pi^{3}\rfloor+...+\lfloor \pi^{10}\rfloor$ without using a calculator. Actually, this is a question which I propose yet fail to solve until now.

More generally, find $\lfloor \pi\rfloor+\lfloor \pi^{2}\rfloor+\lfloor \pi^{3}\rfloor+...+\lfloor \pi^{n}\rfloor$ in terms of $n$.

I hope someone can offer help or suggestions. If you find this interesting just like I do, help me like or re share this note to attract the attention of pros, thank you. I'll sign off here for now.

Cheers!

Note by Donglin Loo
1 year, 11 months ago

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$\pi = 3.14 \implies \lfloor \pi\rfloor= 3$

So, we have to find $\lfloor \pi\rfloor+\lfloor \pi^{2}\rfloor+\lfloor \pi^{3}\rfloor+...+\lfloor \pi^{10}\rfloor$

$\implies 3 + 3^2 + 3^3 + 3^4 + 3^5 .... + 3^{10}$

The above series is in G.P with $a = 3, r = 3, n = 10$

Sum of terms in G.P = $\dfrac{3(3^{10} - 1)}{3 - 1}$

$\implies 1.5(3^{10} - 1)$

Coming to the next part of the question sum of infinite terms in an increasing G.P is infinite.

However, if $n$ is finite the formula would be :

$\dfrac{3(3^n -1)}{3 - 1}$

- 1 year, 11 months ago

@Ram Mohith. You can't assume $\pi=3$ just like that. For instance, $\pi^3=31.000627668$ so $\lfloor \pi^3\rfloor=31$ instead of $3^3=27$. You can verify this with a calculator. The difficult part is that $\pi=3.14159265$ and with exponentiation things can get very random.

- 1 year, 11 months ago

Interesting question, may use this identity $1+\dfrac14+\dfrac19+\dfrac1{16}+\cdots=\dfrac{\pi^2}6$ to do that but I'm not sure...

- 1 year, 11 months ago

Oh wow. I did not think of this. Thanks for your suggestion.

- 1 year, 11 months ago

Take a look at this OEIS ,though it may not be helpful.

- 1 year, 11 months ago

Oh I see. Thanks for your idea

- 1 year, 11 months ago

Perhaps utilize $\dfrac{\pi}{2}=\dfrac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdots}$?

- 1 year, 11 months ago

Wow. Is there a proof for this?

- 1 year, 11 months ago

- 1 year, 11 months ago

@X X Thanks @X X

- 1 year, 11 months ago