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sim0nz12345 · Mar 20, 2007, 02:52 AM

Hi there,
I'm having trouble with this questions...

Could you please give do a step by step solution to this question:

The sum of the first "n" natural numbers is 1/2n(n+1)

How many do you have to add to exceed 1000?

Thanks

Capuchin · Mar 20, 2007, 03:00 AM

so, we form an equation

$\frac{1}{2n(n+1)} = 1000$

Now we solve it for n.

$2n(n+1) = \frac{1}{1000} \\ \\ 2n^2+2n = \frac{1}{1000} \\ \\ 2n^2+2n-\frac{1}{1000} = 0$

You can solve this quadratic however you wish to. You'll need to round your answer up to get the right answer to the question.

sim0nz12345 · Mar 20, 2007, 04:18 AM

Thank-you so much for your answer
It really helps

Capuchin · Mar 20, 2007, 04:25 AM

Sorry, was it $\frac{1}{2n(n+1)}$ or $\frac{1}{2}n(n+1)$ ? - I'm thinking the latter - i didn't check when i wrote the above.

This is why you must be careful of the notation you used for equations, it can be ambiguous.

For the latter:

$\frac{1}{2}n(n+1) = 1000 \\ \\ \frac{1}{2}n^2+\frac{1}{2}n = 1000 \\ \\ \frac{1}{2}n^2+\frac{1}{2}n - 1000 = 0$

and solve in the same way, rounding up.

sim0nz12345 · Mar 20, 2007, 04:42 AM

Yea I knew I typed the first one wrong
Thank you anyway for the help
Much appreciated