Demonstrate that:

$\displaystyle \

\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.

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- Aug 13th 2008, 12:59 PM #1

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- Aug 13th 2008, 01:24 PM #2

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- Aug 13th 2008, 01:42 PM #3
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 $\displaystyle P_{n}$ be the statement that $\displaystyle \left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{n^2 - 1}\right] = \frac{n(n-1)(4n+1)}{6}$.

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

Inductive step: Assume $\displaystyle P_{k}$ to be true. [tex] We must show that $\displaystyle P_{k+1}$ is also true, i.e.

$\displaystyle \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}$$\displaystyle {\color{white}.}\: \: = \: \: \frac{(k+1)(k)(4k+5)}{6}$

Looking at the left hand side (LHS)

$\displaystyle \text{LHS} = \left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{k^{2} + 2k}\right]$

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

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

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

$\displaystyle \text{LHS} = {\color{red}\frac{k(k-1)(4k+1)}{6}} + (2k+1)k$

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

Phew!

- Aug 13th 2008, 03:41 PM #4

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- Aug 14th 2008, 02:49 AM #5

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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:

$\displaystyle \

\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 #6

- Aug 14th 2008, 06:29 AM #7

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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 :

$\displaystyle \

\sum {}

\

$

have a special proprities,for this reason i ask you again:

have ...a special proprities???

for the exemple:

how we can sum that:

$\displaystyle \

\sum\limits_{k = 1}^{n^2 - 1} {\sqrt k } + \sum\limits_{k = 1}^{n^2 - 1} k

\

$