# exer2

• Aug 13th 2008, 12:59 PM
Ortega
exer2
Demonstrate that:
$\
\left[ {\sqrt 1 } \right] + \left[ {\sqrt 2 } \right] + \left[ {\sqrt 3 } \right] + ... + \left[ {\sqrt {n^2 - 1} } \right] = \frac{{n(n - 1)(4p + 1)}}{6}
\
$

pls solve it with explanation.
• Aug 13th 2008, 01:24 PM
TKHunny
Either I don't know what "p" is or that just makes no sense. That's an awful lot of unrelated irrational numbers magically adding up to a Rational.

There must be a typo or simply proving it incorrect is easy enough.
• Aug 13th 2008, 01:42 PM
o_O
What's that p doing there .. I'm guessing it should be an 'n'. In any case, this sounds like a proof by induction.

Let $P_{n}$ be the statement that $\left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{n^2 - 1}\right] = \frac{n(n-1)(4n+1)}{6}$.

Base case: P(1): 0 = 0. True.

Inductive step: Assume $P_{k}$ to be true. [tex] We must show that $P_{k+1}$ is also true, i.e.
$\left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{(k+1)^2 - 1}\right] = \frac{(k+1)((k+1)-1)(4(k+1)+1)}{6}$ ${\color{white}.}\: \: = \: \: \frac{(k+1)(k)(4k+5)}{6}$

Looking at the left hand side (LHS)
$\text{LHS} = \left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{k^{2} + 2k}\right]$
$\text{LHS} = {\color{red}\left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{k^2 - 1}\right] +}$ $\: \left[\sqrt{k^2}\right] + \left[\sqrt{k^2 + 1}\right] + ... + \left[\sqrt{k^2 + 2k - 1}\right] + \left[\sqrt{k^2 + 2k}\right]$

We see that $\left[\sqrt{k^2} \right] = k$ and the next integer up would be $\left[\sqrt{(k+1)^2}\right] = \left[\sqrt{k^2 + 2k + 1}\right] = k + 1$. So, since our last term is $k^2 + 2k$ (one less), then we see that : $\left[\sqrt{k^2}\right] = \left[\sqrt{k^2 + 1}\right] = ... = \left[\sqrt{k^2 + 2k}\right] = k$

Now, rewriting the equation and noticing that what is in red is $P_{k}$

$\text{LHS} = {\color{red}\frac{k(k-1)(4k+1)}{6}} + \underbrace{k + k + ... + k}_{\text{(2k + 1) times}}$
$\text{LHS} = {\color{red}\frac{k(k-1)(4k+1)}{6}} + (2k+1)k$

Now hopefully if you simplify it, you'll get what you want.

Phew!
• Aug 13th 2008, 03:41 PM
TKHunny
Ah! The integer function strikes again. I wish that notation were clear or standard.
• Aug 14th 2008, 02:49 AM
Ortega
Excuse me it's "n" not "p".
I have one demnostration that not need induction but im not under stand it.i will post the solution and i expect you to explain to me:
the demonstration become:
$\
\sum\limits_{k = 1}^{n^2 - 1} {\left[ {\sqrt k } \right]} = \sum\limits_{k = 1}^{n - 1} {k\left[ {(k + 1)^2 - k^2 } \right]} = \sum\limits_{k = 1}^{n - 1} {2k^2 + k = ....}
\
$

Im not understand why???can you explain to me???
• Aug 14th 2008, 04:27 AM
Plato
Quote:

Originally Posted by o_O
We see that $\left[\sqrt{k^2} \right] = k$ and the next integer up would be $\left[\sqrt{(k+1)^2}\right] = \left[\sqrt{k^2 + 2k + 1}\right] = k + 1$. So, since our last term is $k^2 + 2k$ (one less), then we see that : $\underbrace{\left[\sqrt{k^2}\right] = \left[\sqrt{k^2 + 1}\right] = ... = \left[\sqrt{k^2 + 2k}\right] = k}$

Quote:

Originally Posted by Ortega
$\sum\limits_{k = 1}^{n^2 - 1} {\left[ {\sqrt k } \right]} = \sum\limits_{k = 1}^{n - 1} {k\left[ {(k + 1)^2 - k^2 } \right]} = \sum\limits_{k = 1}^{n - 1} {2k^2 + k = ....}$
Im not understand why???can you explain to me???

That is the explanation!
• Aug 14th 2008, 06:29 AM
Ortega
Quote:

Originally Posted by Plato
That is the explanation!

yes i undetsand what he wrote.it's more simple than :
.....(what i wrote)
But in my demonstration(in the book)they are not precise anything of that
they write directly the solution .
i thought that :
$\
\sum {}
\
$

have a special proprities,for this reason i ask you again:
have ...a special proprities???
for the exemple:
how we can sum that:
$\
\sum\limits_{k = 1}^{n^2 - 1} {\sqrt k } + \sum\limits_{k = 1}^{n^2 - 1} k
\
$