1 +2+3+4...+999998+999999+1000000 =

[Bronium 512]

1 +2+3+4...+999998+999999+1000000 = 500000500000

So....Am I right?

Edited August 21, 2012 by Bronium

[Azura 1,107]

No, it's 3,000,008.

 

I have no idea how you got that number..

[Bronium 512]

No, you misunderstand me. Add every number from 1 to 1000000. That's what the ... means.

 

Sorry, I accidentally wrote 999999 twice when it should've been 999998 + 999999

[Ashbad 969]

1 +2+3+4...+999999+999999+1000000 = 500000500000

So....Am I right?

 

#include <stdio.h>
#include <stdlib.h>

int main()
{
    unsigned long long int acc = 0;
    for(int i = 1; i <= 1000000; i++) {
        acc += i;
    }
    printf("%llu", acc);
    return 0;
}

500000500000 is what I'm getting, so seems you are indeed correct.

[Bronium 512]

Huzzah! Oh and the equation I got was (n+1)*(0.5n).

Where n is the number you want to add up to. So far n is an element of N. Not sure of others just yet. Will update.

 

Huzzah! Oh and the equation I got was (n+1)*(0.5n).

Where n is the number you want to add up to. So far n is an element of N. Not sure of others just yet. Will update.

[Ashbad 969]

Huzzah! Oh and the equation I got was (n+1)*(0.5n).

Where n is the number you want to add up to. So far n is an element of N. Not sure of others just yet. Will update.

 

If you want to solve these sorts of equations the non-fun way (as in, code aside), it's a simple sigma function, where i = 1 and n = 1000000. The formula for finding the summation is n(n+1)/2. Plug in your numbers:

 

1000000(1000001)/2 -> 500000500000

[Bronium 512]

I wonder if (n+1)*(0.5n) works for negative numbers. Or is it just natural numbers?

 

And this is the fun way ;.;

[Ashbad 969]

I wonder if (n+1)*(0.5n) works for negative numbers. Or is it just natural numbers?

 

And this is the fun way ;.;

 

Only works when i = 1. Otherwise, there are other equations for summation that you need to use. http://en.wikipedia.org/wiki/Summation

[Bronium 512]

Great, send me to wikipedia at 12am at night. And it's about maths. Dammit.

[tanoshii 131]

> sum[1..100000]
=> 5000050000

Haskell

[thegoodhen 698]

I'm sorry, but this trick has been actually found decades ago by one guy, not sure who... Fibonacci? Hm, something like that-he was a kid and their teacher told em to sum all nums up to 100.

 

unsigned long long int acc = 0;

 

 

unsigned long long int? "Mann learnt nie aus" as German ppl say-Basically, I just learned this is possible xD

[Ashbad 969]

unsigned long long int? "Mann learnt nie aus" as German ppl say-Basically, I just learned this is possible xD

 

Means to use a long of an integer as natively possible though if you're feeling ballsy, GCC suppports 128-bit integers as __int128.

 

> sum[1..100000]
=> 5000050000

Haskell

 

Biggest toy language ever made

[ProjectRKA 2,848]

So, everypony... what is 5 times 2 ? Come on now, don't be shy :3

[Bronium 512]

It was Carl Gauss, I think. He counted up to a hundred and I wanted to make sure I could use his trick. I was just making sure that I was right.