# Thread: Square Root of Sum of the Numbers

1. ## Square Root of Sum of the Numbers

What is the square root of the sum of the first 2007 positive odd integers?

2. Originally Posted by Rimas
What is the square root of the sum of the first 2007 positive odd integers?

$\sqrt{\sum^{2007}_{n=1}(2n-1)}$

$=\left(2\sum^{2007}_{n=1}(n)-\sum^{2007}_{n=1}(1)\right)^\frac{1}{2}$

$=\left(2\left(\frac{2007(2007+1)}{2}\right)-(2007)\right)^\frac{1}{2}$

$=\left(2007(2007+1-1)\right)^\frac{1}{2}$

$=\left((2007)^2\right)^\frac{1}{2}$

$=2007$

3. Hello, Rimas!

This could be a trick question . . .

What is the square root of the sum of the first 2007 positive odd integers?
We have: . $S \;= \;\underbrace{1 + 3 + 5 + 7 + \cdots }_{2007\text{ terms}}$

This is an arithmetic series with first term $a = 1$ and common difference $d = 2$

The sum of the first $n$ terms is: . $\frac{n}{2}[2a + (n-1)d]$

So we have: . $S \;=\;\frac{2007}{2}[2\cdot1 + 2006(2)] \;=\;2007^2$

Therefore: . $\sqrt{S}\;=\;\sqrt{2007^2} \;=\;2007$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you knew the following bit of mathematical trivia,
. . you could have answered it immediately.

$\begin{array}{ccccc}1 & = & 1 & = & 1^2 \\1 + 3 & = & 4 & = & 2^2 \\1 + 3 + 5 & = & 9 & = & 3^2 \\1 + 3 + 5 + 7 & = & 16 & = & 4^2 \\1 + 3 + 5 + 7 + 9 & = & 25 & = & 5^2\end{array}$

Get it?
The sum of the first $n$ odd number is always $n^2$.

So the sum of the first 2007 odd number is $2007^2$.

And its square root is, of course, $2007$.

4. Originally Posted by Soroban
Hello, Rimas!

This could be a trick question . . .

We have: . $S \;= \;\underbrace{1 + 3 + 5 + 7 + \cdots }_{2007\text{ terms}}$

This is an arithmetic series with first term $a = 1$ and common difference $d = 2$

The sum of the first $n$ terms is: . $\frac{n}{2}[2a + (n-1)d]$

So we have: . $S \;=\;\frac{2007}{2}[2\cdot1 + 2006(2)] \;=\;2007^2$

Therefore: . $\sqrt{S}\;=\;\sqrt{2007^2} \;=\;2007$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you knew the following bit of mathematical trivia,
. . you could have answered it immediately.

$\begin{array}{ccccc}1 & = & 1 & = & 1^2 \\1 + 3 & = & 4 & = & 2^2 \\1 + 3 + 5 & = & 9 & = & 3^2 \\1 + 3 + 5 + 7 & = & 16 & = & 4^2 \\1 + 3 + 5 + 7 + 9 & = & 25 & = & 5^2\end{array}$

Get it?
The sum of the first $n$ odd number is always $n^2$.

So the sum of the first 2007 odd number is $2007^2$.

And its square root is, of course, $2007$.

Thanks Soroban. I was thinking of doing it ecMathGeeks way, but never considered arithmetic series. That trivia was nice as well

5. Hello, ecMathGeek!

A straight-forward, no-nonsense approach . . . great job!

I intended to use your method, but something bothered me.

The square root of the sum? . . . Won't that be some ugly decimal?
. . Why in the world would they want such a . . . Oh!

I remembered that bit of trivia . . . and I was literally LOL.

Then I had to devise an approach . . . and I used Arithmetic Series.

6. Originally Posted by Soroban
Hello, ecMathGeek!

A straight-forward, no-nonsense approach . . . great job!

I intended to use your method, but something bothered me.

The square root of the sum? . . . Won't that be some ugly decimal?
. . Why in the world would they want such a . . . Oh!

I remembered that bit of trivia . . . and I was literally LOL.

Then I had to devise an approach . . . and I used Arithmetic Series.

Thank you.

I started it thinking the same thing (but figured/hoped it would work out in the end). Just after I set up the summation, I remembered that same trivia that you spoke of and knew at that point what the answer would be. So, relieved that there would be a solution, I persevered until I reached it.