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hi, I have 2 questions here that I am stuck on. I thought I may as well stick them in the same topic to avoid cluttering the forum. hope you can help, thnx

Fred invests an amount of money in an account paying

Find a formula for

where

[tex]2m = m(\frac{100+r}{100})^{n}[/tex]

[tex]2 = (\frac{100+r}{100})^{n}[/tex]

[tex]\sqrt[n]{2} = \frac{100+r}{100}[/tex]

[tex]100\sqrt[n]{2} = 100 + r[/tex]

[tex]r = 100\sqrt[n]{2} - 100[/tex]

now, that seems to be the right answer but looks kinda ugly... first of all, is that answer right? and secondly, if so, is there a nicer way to put it :S ?

thnx

Using this result, or otherwise

[tex](m^2+1)(n^2+1) = (m + n)^2 + (mn -1)^2[/tex]

write 500050 as the sum of 2 square numbers.

I havn't been told how to do this. maybe because this is on a past paper it isn't on the syllabus anymore, but still, I'd like to know how to do it because i doubt its that hard once you know how.

thnx

**QUESTION 1**## Homework Statement

Fred invests an amount of money in an account paying

*r*% compound interest per annum. The amount of money doubles after*n*years.Find a formula for

*r*in terms of*n*.## Homework Equations

## The Attempt at a Solution

where

*m*is the initial money][tex]2m = m(\frac{100+r}{100})^{n}[/tex]

[tex]2 = (\frac{100+r}{100})^{n}[/tex]

[tex]\sqrt[n]{2} = \frac{100+r}{100}[/tex]

[tex]100\sqrt[n]{2} = 100 + r[/tex]

[tex]r = 100\sqrt[n]{2} - 100[/tex]

now, that seems to be the right answer but looks kinda ugly... first of all, is that answer right? and secondly, if so, is there a nicer way to put it :S ?

thnx

**QUESTION 2**## Homework Statement

Using this result, or otherwise

*I proved for the previous question that*[tex](m^2+1)(n^2+1) = (m + n)^2 + (mn -1)^2[/tex]

write 500050 as the sum of 2 square numbers.

## Homework Equations

## The Attempt at a Solution

I havn't been told how to do this. maybe because this is on a past paper it isn't on the syllabus anymore, but still, I'd like to know how to do it because i doubt its that hard once you know how.

thnx

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