Question

Find the sum of the first thousand positive integer the first n positive integers

Answer 1

1000*(1001)/2 is it okk...by solving that u may get easily answer

Answer 2

Here's a proof that uses concepts from combinatorics :

Suppose we have [math](n+1)[/math] points in space:

[math]\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot [/math]

Now , let's see in how many ways we can choose any two points from this set of points:

Take a look at the leftmost point. If we were to include this point in our choice of two points, we could choose any one of the other [math]n[/math] points that we have on our list to go with our chosen leftmost point.

So, now we have [math]n[/math] choices in our hand.

In a similar fashion, if we were to include the point second from the left in our choice of two points , we would have [math](n-1)[/math] choices . (Note that we can't go left from this chosen point because that has already been taken into account previously )

Now, you can probably guess that we can have [math](n-2)[/math] choices if we wish to include the point second from the left in our choice of two points.

And so on.......

Obvoiously, you can see that the number of ways we can choose any two points from [math](n+1)[/math] points in space is : [math]{ n+ (n-1) + (n-2) + ....+3+2+1}[/math]

or [math]{1+2+3+....+n}[/math]

Now, if you can recall , using combinatorics we can again say that the number of ways we can choose any two points from [math](n+1)[/math] points in space is [math]^{(n+1)}[/math]C[math]_2[/math] or [math]\frac{n(n+1)}{2}[/math]

Therefore ,

[math]1+2+3+.....n = \frac{n(n+1)}{2}[/math]

[math]\fbox{Proved}[/math]