1. ## Math program

I don't understand why my program doesn't work... Its supposed to find the sum of:

1*2 + 3*4 + 5*6 ... + 2003*2004
When i output what is being held in array[0] it gives me like 4.022+06 or something whereas it should be 1 and array[1] should be two

Code:
```#include <iostream.h>
#include <math.h>

void main() {
int i, j;
double array[1002];
double sum=0;
double square;

for (j=0; j<1002; j++) {
for (i=1; i<=2004; i+=2) {
array[j]=i*(i+1);
}
}

for (i=0; i<1002; i++) {
sum+=array[i];
}
cout<<array[0]<<endl;
cout<<sum<<endl;
}```

2. Your inner loop does nothing - except to put 2004 * 2005 into every array element

3. Oh yeah you're right thanks a lot

4. k i fixed it

Code:
```	for (j=1; j<=1002; j++) {
array[j-1]=(2*j)*(2*j -1);
}```

5. Code:
```for (j=0; j<1002; j++) {
for (i=1; i<=2004; i+=2) {
array[j]+=i*(i+1);
}
}```
would this work note you must init the array to zero

6. It would work but why have an array with 1002 elements, that are all the same?

7. Code:
```int sum(int maxNumber)
{
if (maxNumber<=2) return maxNumber;
return maxNumber*(maxNumber-1)+sum(maxNumber-2);
}```
Thought I'd try a recursive solution.
Is this correct. Not very pretty thats for sure.

8. well my new and improved code sets

array [0]: 1*2
array [1]: 3*4
array [2]: 5*6

...

array [1002] 2003*2004

9. Been thinking about your new and improved code. . . I think that the last entry is wrong.

Code:
`array[1001]: 2003*2004`
Reasoning: each "n" entry of the array should be equal to ([(n+1)*2]-1)*[(n+1)*2]

just2peachy

P.S.
if you look closely at these numbers:
array[0]=1*2=2(1)
array[1]=3*4=2(6)
array[2]=5*6=2(15)
array[3]=7*8=2(28)
.
.
.
Examine the pattern forming. Do you see a special number pattern with 1, 6, 15, 28, . . .

If you've messed around with triangular numbers or hexagonal numbers, you should see some bells and whistles.

Try a google search with triangular numbers. You'll find that each
T(2n+1) or each odd triangular number fitts this pattern. There are many formulas for dealing with the summation of triangular numbers or in this case hexagonal numbers. The summation of these numbers could probably be found with one quick formula.