Q:

Nested loops result

I really don't know how to find out the result of nested loops. 
For example in the following pseudo-code, I can't sort out what will be given at the end of execution. I'll be so glad if anyone gives me a simple solution.
r <- 0
for i <- 1 to n do 
  for j <- 1 to i do
    for k <- j to i+j do
      r <- r + 1
return r

Question is: 
What is the result of the code and give the result r in terms of n? 
I write it but every time I get confused. 

A:

In your pseudo-code, Inner most loop, k <- j to i+j can be written as k <- 0 to i (this is by removing j). Hence your code can be simplified as follows: 
r <- 0
for i <- 1 to n do 
    for j <- 1 to i do
      for k <- 0 to i do   // notice here `j` removed
        r <- r + 1
return r

Based on this pseudo-code, I have written a C program(as below) to generate sequence for N = 1 to 10. (you originally tagged question as java but I am writing c code because what you wants is independent of language constraints)
#include<stdio.h>
int main(){
  int i =0, k =0, j =0, n =0;
  int N =0; 
  int r =0;
  N =10;
  for (n=1; n <= N; n++){
  // unindented code here
  r =0;
  for (i=1; i<=n; i++)
      for (j=1; j<=i; j++)
          for (k=0; k<=i; k++)
              r++;    
  printf("\n N=%d  result = %d",n, r); 
  }
  printf("\n");
}

Output of this program is something like:
 $ ./a.out     
 N=1  result = 2
 N=2  result = 8
 N=3  result = 20
 N=4  result = 40
 N=5  result = 70
 N=6  result = 112
 N=7  result = 168
 N=8  result = 240
 N=9  result = 330
 N=10  result = 440

Then, Tried to explore, How does it work? with some diagrams:  
Execution Tree For N=1:    
1<=i<=1,              (i=1)
                        |
1<=j<=i,              (j=1)
                      /   \
0<=k<=i,          (K=0)    (K=1)
                   |        |  
r=0                r++      r++    => r = 2
                 ( 1   +    1 )

That is     (1*2) = 2 
Tree For N=2:
1<=i<=2,         (i=1)-----------------------(i=2)
                  |                 |---------|------|
1<=j<=i,        (j=1)           (j=1)              (j=2)
                /   \          /  |  \            /  |  \
0<=k<=i,    (K=0)    (K=1)  (K=0)(K=1)(k=2)    (K=0)(K=1)(k=2)
              |        |      |     |    |       |     |    |
r=0          r++      r++   r++   r++  r++     r++   r++  r++    => 8
            --------------  ---------------------------------
            ( 1   +    1)   (     3         +      3        )

That is (1 + 1) + (3 + 3) = 8
Similarly I drawn a tree for N=3:
1<=i<=3,        (i=1)-----------------------(i=2)--------------------------------------------(i=3)
                  |                 |---------|------|                  |----------------------|----------------------|
1<=j<=3,        (j=1)             (j=1)            (j=2)            (   j=1   )           (   j=2    )           (   j=3    )
                /   \            /  |  \          /  |  \          / |       |  \        / |       |  \          / |       | \
0<=k<=i,    (K=0)   (K=1)    (K=0)(K=1)(k=2)   (K=0)(K=1)(k=2)    /  |       |   \      /  |       |   \        /  |       |  \
              |       |        |    |    |       |    |    |     (K=0)(K=1)(k=2)(k=3)  (K=0)(K=1)(k=2)(k=3)  (K=0)(K=1)(k=2)(k=3)
r=0          r++     r++      r++  r++  r++     r++  r++  r++     r++  r++   r++  r++   r++  r++  r++  r++   r++  r++   r++   r++

That is (1 + 1) + (3 + 3) + (4 + 4+ 4)= 20
N = 1,   (1 + 1) = 2  
N = 2,   (1 + 1) + (3 + 3) = 8  
N = 3,   (1 + 1) + (3 + 3) + (4 + 4 + 4)= 20  
N = 4,   (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5)  =  40  
N = 5,   (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5) + (6 + 6 + 6 + 6 + 6) = 70  
N = 6,   (1 + 1) + (3 + 3) + (4 + 4 + 4) + (5 + 5 + 5 + 5) + (6 + 6 + 6 + 6 + 6) + (7 + 7 + 7 + 7 + 7 + 7)= 112  

For  N=6 we can also be write above sequence as:
(1*2) + (2*3) + (3*4) + (4*5) + (5*6) + (6*7)    

Finally, I could understand that sum of N in three loop is:
(1*2) + (2*3) + (3*4) + (4*5) + (5*6) + ... + (N * (N+1))

With help from math.stackexchange.com, I could simplify this equation:
I asked here: How to simplify summation equation in terms of N? 

 

So as I commented to your question, Result in term of N is ( ((N) * (N+1) * (N+2)) / 3 ).
And, I think its correct. I cross checked it as follows:
N = 1,    (1 * 2 * 3)/3  = 2

N = 2,    (2 * 3 * 4)/3 =  8

N = 3,    (3 * 4 * 5)/3 =  20

N = 4,    (4 * 5 * 6)/3 =  40

N = 5,    (5 * 6 * 7)/3 =  70    

