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    <p>In your pseudo-code, Inner most loop, <code>k &lt;- j to i+j</code> can be written as <code>k &lt;- 0 to i</code> (<em>this is by removing <code>j</code></em>). Hence your code can be simplified as follows: </p> <pre><code>r &lt;- 0 for i &lt;- 1 to n do for j &lt;- 1 to i do for k &lt;- 0 to i do // notice here `j` removed r &lt;- r + 1 return r </code></pre> <p>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 <a href="/questions/tagged/java" class="post-tag" title="show questions tagged &#39;java&#39;" rel="tag">java</a> but I am writing <a href="/questions/tagged/c" class="post-tag" title="show questions tagged &#39;c&#39;" rel="tag">c</a> code because what you wants is independent of language constraints)</p> <pre><code>#include&lt;stdio.h&gt; int main(){ int i =0, k =0, j =0, n =0; int N =0; int r =0; N =10; for (n=1; n &lt;= N; n++){ // unindented code here r =0; for (i=1; i&lt;=n; i++) for (j=1; j&lt;=i; j++) for (k=0; k&lt;=i; k++) r++; printf("\n N=%d result = %d",n, r); } printf("\n"); } </code></pre> <p>Output of this program is something like:</p> <pre><code> $ ./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 </code></pre> <p>Then, Tried to explore, <strong>How does it work?</strong> with some diagrams: </p> <p>Execution Tree For <code>N=1</code>: </p> <pre><code>1&lt;=i&lt;=1, (i=1) | 1&lt;=j&lt;=i, (j=1) / \ 0&lt;=k&lt;=i, (K=0) (K=1) | | r=0 r++ r++ =&gt; r = 2 ( 1 + 1 ) </code></pre> <p>That is <code>(1*2) = 2</code> </p> <p>Tree For <code>N=2</code>:</p> <pre><code>1&lt;=i&lt;=2, (i=1)-----------------------(i=2) | |---------|------| 1&lt;=j&lt;=i, (j=1) (j=1) (j=2) / \ / | \ / | \ 0&lt;=k&lt;=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++ =&gt; 8 -------------- --------------------------------- ( 1 + 1) ( 3 + 3 ) </code></pre> <p>That is <code>(1 + 1) + (3 + 3) = 8</code></p> <p>Similarly I drawn a tree for <code>N=3</code>:</p> <pre><code>1&lt;=i&lt;=3, (i=1)-----------------------(i=2)--------------------------------------------(i=3) | |---------|------| |----------------------|----------------------| 1&lt;=j&lt;=3, (j=1) (j=1) (j=2) ( j=1 ) ( j=2 ) ( j=3 ) / \ / | \ / | \ / | | \ / | | \ / | | \ 0&lt;=k&lt;=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++ </code></pre> <p>That is <code>(1 + 1) + (3 + 3) + (4 + 4+ 4)= 20</code></p> <pre><code>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 </code></pre> <p>For N=6 we can also be write above sequence as:</p> <pre><code>(1*2) + (2*3) + (3*4) + (4*5) + (5*6) + (6*7) </code></pre> <p>Finally, I could understand that sum of <code>N</code> in three loop is:</p> <pre><code>(1*2) + (2*3) + (3*4) + (4*5) + (5*6) + ... + (N * (N+1)) </code></pre> <p>With help from math.stackexchange.com, I could simplify this equation:<br> I asked here: <a href="https://math.stackexchange.com/questions/416322/how-to-simplify-summation-equation-in-terms-of-n">How to simplify summation equation in terms of N?</a> </p> <blockquote> <p><img src="https://i.stack.imgur.com/CdzaI.png" alt="image"> </p> </blockquote> <p>So as I commented to your question, Result in term of N is <code>( ((N) * (N+1) * (N+2)) / 3 )</code>.<br> And, I think its correct. I cross checked it as follows:</p> <pre><code>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 </code></pre>
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    1. COcan I just don't replace k <- j to i+j by k <- 0 to i?
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    2. COalso I don't get how did u come up with the equation (1+1)+(3+3)...
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    3. CO@NavidKoochooloo Yes I replaced `k <- j to i+j` by `k <- 0 to i`,
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