View Full Version : Sum of the first natural numbers
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