Sum of pairwise products

For given any unsigned int n find the final value of ∑(i*j) where (1<=i<=n) and (i <= j <= n).
Examples:

Input : n = 3
Output : 25
We get the sum as following. Note that
first term i varies from 1 to 3 and second
term values from value of first term to n.
1*1 + 1*2 + 1*3 + 2*2 + 2*3 + 3*3 = 25

Input : 5
Output : 140

Method 1 (Simple) We run two loops and compute the required sum. 

C++

// Simple CPP program to find sum
// of given series.
#include <bits/stdc++.h>
using namespace std;

long long int findSum(int n)
{
   long long int sum = 0;
   for (int i=1; i<=n; i++)
     for (int j=i; j<=n; j++)
        sum = sum + i*j;
   return sum;
}

int main()
{
    int n = 5;
    cout << findSum(n);
    return 0;
}

Java

// Simple Java program to find sum
// of given series.
class GFG {

    static int findSum(int n)
    {
        int sum = 0;

        for (int i=1; i<=n; i++)
            for (int j=i; j<=n; j++)
                sum = sum + i*j;

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {

        int n = 5;

        System.out.println(findSum(n));
    }
}

// This code is contributed by Smitha Dinesh Semwal.

Python3

# Simple Python3 program to
# find sum of given series.

def findSum(n) :
    sm = 0
    for i in range(1, n + 1) :
        for j in range(i, n + 1) :
            sm = sm + i * j

    return sm

# Driver Code
n = 5
print(findSum(n))

# This code is contributed by Nikita Tiwari.

C#

// Simple C# program to find sum
// of given series.
class GFG {

    static int findSum(int n)
    {
        int sum = 0;

        for (int i=1; i<=n; i++)
            for (int j=i; j<=n; j++)
                sum = sum + i*j;

        return sum;
    }

    // Driver code
    public static void Main()
    {

        int n = 5;

        System.Console.WriteLine(findSum(n));
    }
}

// This code is contributed by mits.

PHP

<?php
// Simple PHP program to find sum
// of given series.

function findSum($n)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
        for ($j = $i; $j <= $n; $j++)
            $sum = $sum + $i * $j;
    return $sum;
}

// Driver Code
$n = 5;
echo findSum($n);

// This code is contributed by mits
?>

Javascript

<script>

// Simple JavaScript program to find sum
// of given series.

    function findSum(n)
    {
        let sum = 0;

        for (let i=1; i<=n; i++)
            for (let j=i; j<=n; j++)
                sum = sum + i*j;

        return sum;
    }

// Driver Code

        let n = 5;

        document.write(findSum(n));

</script>

Output:

140

Time Complexity: O(n^2).
Method 2 (Better)
We can observe following in the given problem. 
1 is multiplied with all numbers from 1 to n. 
2 is multiplied with all numbers from 2 to n. 
……………………………………… 
……………………………………… 
i is multiplied with all numbers from i to n.
We compute sum of first n natural numbers which is our first term. For remaining terms, we multiply i with sum of numbers from i to n. We keep track of this sum by subtracting i from initial sum in every iteration.

C++

// Efficient CPP program to find sum
// of given series.
#include <bits/stdc++.h>
using namespace std;

long long int findSum(int n)
{
   long long int multiTerms = n * (n + 1) / 2;

   // Sum of multiples of 1 is 1 * (1 + 2 + ..)
   long long int sum = multiTerms;

   // Adding sum of multiples of numbers other
   // than 1, starting from 2.
   for (int i=2; i<=n; i++)
   {
       // Subtract previous number
       // from current multiple.
       multiTerms = multiTerms - (i - 1);

       // For example, for 2, we get sum
       // as (2 + 3 + 4 + ....) * 2
       sum = sum + multiTerms * i;
   }
   return sum;
}

// Driver code
int main()
{
    int n = 5;
    cout << findSum(n);
    return 0;
}

Java

// Efficient Java program to find sum
// of given series.
class GFG {

    static int findSum(int n)
    {

        int multiTerms = n * (n + 1) / 2;

        // Sum of multiples of 1 is 1 * (1 + 2 + ..)
        int sum = multiTerms;

        // Adding sum of multiples of numbers other
        // than 1, starting from 2.
        for (int i = 2; i <= n; i++)
        {

            // Subtract previous number
            // from current multiple.
            multiTerms = multiTerms - (i - 1);

            // For example, for 2, we get sum
            // as (2 + 3 + 4 + ....) * 2
            sum = sum + multiTerms*i;
        }

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {

        int n = 5;

        System.out.println(findSum(n));
    }
}

// This code is contributed by Smitha Dinesh Semwal.

Python3

# Efficient Python3 program
# to find sum of given series.

def findSum(n) :
    multiTerms = n * (n + 1) // 2

    # Sum of multiples of 1 is 1 * (1 + 2 + ..)
    sm = multiTerms

    # Adding sum of multiples of numbers
    # other than 1, starting from 2.
    for i in range(2, n+1) :

        # Subtract previous number
        # from current multiple.
        multiTerms = multiTerms - (i - 1)

        # For example, for 2, we get sum
        # as (2 + 3 + 4 + ....) * 2
        sm = sm + multiTerms * i

    return sm

# Driver code
n = 5
print(findSum(n))

# This code is contributed by Nikita Tiwari.

C#

// C# program to find sum
// of given series.
using System;
class GFG {

    static int findSum(int n)
    {

        int multiTerms = n * (n + 1) / 2;

        // Sum of multiples of 1 is 1 * (1 + 2 + ..)
        int sum = multiTerms;

        // Adding sum of multiples of numbers other
        // than 1, starting from 2.
        for (int i = 2; i <= n; i++)
        {

            // Subtract previous number
            // from current multiple.
            multiTerms = multiTerms - (i - 1);

            // For example, for 2, we get sum
            // as (2 + 3 + 4 + ....) * 2
            sum = sum + multiTerms*i;
        }

        return sum;
    }

    // Driver code
    public static void Main()
    {

        int n = 5;

        Console.WriteLine(findSum(n));
    }
}

// This code is contributed by Mukul Singh.

PHP

<?php
// Efficient PHP program to find sum
// of given series.

function findSum($n)
{
    $multiTerms = (int)($n * ($n + 1) / 2);

// Sum of multiples of 1 is 1 * (1 + 2 + ..)
$sum = $multiTerms;

// Adding sum of multiples of numbers other
// than 1, starting from 2.
for ($i=2; $i<=$n; $i++)
{
    // Subtract previous number
    // from current multiple.
    $multiTerms = $multiTerms - ($i - 1);

    // For example, for 2, we get sum
    // as (2 + 3 + 4 + ....) * 2
    $sum = $sum + $multiTerms * $i;
}
return $sum;
}

// Driver code

    $n = 5;
    echo findSum($n);

//This code is contributed by mits
?>

Javascript

<script>

    // Javascript program to find
    // sum of given series.

    function findSum(n)
    {

        let multiTerms = n * (n + 1) / 2;

        // Sum of multiples of 1 is 1 * (1 + 2 + ..)
        let sum = multiTerms;

        // Adding sum of multiples
        // of numbers other
        // than 1, starting from 2.
        for (let i = 2; i <= n; i++)
        {

            // Subtract previous number
            // from current multiple.
            multiTerms = multiTerms - (i - 1);

            // For example, for 2, we get sum
            // as (2 + 3 + 4 + ....) * 2
            sum = sum + multiTerms*i;
        }

        return sum;
    }

    let n = 5;

    document.write(findSum(n));

</script>

Output:

140

Time Complexity: O(n).
Method 3 (Efficient)
The whole calculation can be done in the O(1) time as there is a closed-form expression for the sum. Let i and j run from 1 through n. We want to compute S = sum(i*j) overall i and j such that i <= j otherwise we would have duplicates such as 2*3 + …+3*2; on the other hand, having i = j is OK which gives us 2*2 + … This is all by the problem specification. (Sorry if my notation is bizarre.)
Now, there are two kinds of terms in the sum S: those with i = j (squares, denoted S1) and those with i j always, but the value is the same, by symmetry: 2 * S2 = (sum i)^2 – sum (i^2) . Note that 2*S2 = S3 – S1 as the latter sum is just S1; the former sum (denoted S3) is new here, but there is a closed-form solution for it too. We can now eliminate the mixed term completely: S = S1 + S2 = (S1 + S3) / 2.
Since sum(i) = n*(n+1)/2, we get S3 = n*n*(n+1)*(n+1)/4 ; likewise for the sum of squares: S1 = n*(2*n+1)*(n+1)/6. The final expression simplifies to: 
S = n*(n+1)*(n+2)*(3*n+1)/24 . (As an exercise, you may want to prove that the numerator is indeed divisible by 24.) 

C++

// Efficient CPP program to find sum
// of given series.
#include <bits/stdc++.h>
using namespace std;

long long int findSum(int n)
{
   return n*(n+1)*(n+2)*(3*n+1)/24;
}

// Driver code
int main()
{
    int n = 5;
    cout << findSum(n);
    return 0;
}

Java

// Efficient Java program to find sum
// of given series.
class GFG {

    static int findSum(int n)
    {
        return n * (n + 1) * (n + 2) *
                                 (3 * n + 1) / 24;
    }

    // Driver code
    public static void main(String[] args)
    {

        int n = 5;

        System.out.println(findSum(n));
    }
}

// This code is contributed by Smitha Dinesh Semwal.

Python3

# Efficient Python3 program to find
# sum of given series.

def findSum(n):

    return n * (n + 1) * (n + 2) * (3 * n + 1) / 24

# Driver code
n = 5
print(int(findSum(n)))

# This code is contributed by Smitha Dinesh Semwal.

C#

// Efficient C# program
// to find sum of given
// series.
using System;

class GFG
{
static int findSum(int n)
{
    return n * (n + 1) * (n + 2) *
                 (3 * n + 1) / 24;
}

// Driver code
static public void Main ()
{
    int n = 5;

    Console.WriteLine(findSum(n));
}
}

// This code is contributed
// by ajit.

PHP

<?php
// Efficient PHP
// program to find sum
// of given series.

function findSum($n)
{
    return $n * ($n + 1) *
           ($n + 2) * (3 *
           $n + 1) / 24;
}

// Driver code
$n = 5;
echo findSum($n);

// This code is contributed
// by akt_mit
?>

Javascript

<script>

    // Efficient Javascript program
    // to find sum of given
    // series.

    function findSum(n)
    {
        return n * (n + 1) * (n + 2) * (3 * n + 1) / 24;
    }

    let n = 5;

    document.write(findSum(n));

</script>

Output:

140

Time Complexity: O(1).
Thanks to diprey1 for suggesting this efficient solution.