Find the sum of elements of the Matrix generated by the given rules

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Given three integers A, B, and R, the task is to find the sum of all the elements of the matrix generated by the given rules:

The first row will contain a single element which is A and the rest of the elements will be 0.
The next row will contain two elements all of which are (A + B) and the rest are 0s.
Third row will contain (A + B + B) three times and the rest are 0s and so on.
The matrix will contain only R rows.

For example, if A = 5, B = 3 and R = 3 then the matrix will be:
5 0 0
8 8 0
11 11 11
Examples:

Input: A = 5, B = 3, R = 3
Output: 54
5 + 8 + 8 + 11 + 11 + 11 = 54
Input: A = 7, B = 56, R = 1
Output: 7

Approach: Initialize sum = 0 and for every 1 ? i ? R update sum = sum + (i * A). After every iteration update A = A + B. Print the final sum in the end.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the required sum
int sum(int A, int B, int R)
{
 
    // To store the sum
    int sum = 0;
 
    // For every row
    for (int i = 1; i <= R; i++) {
 
        // Update the sum as A appears i number
        // of times in the current row
        sum = sum + (i * A);
 
        // Update A for the next row
        A = A + B;
    }
 
    // Return the sum
    return sum;
}
 
// Driver code
int main()
{
 
    int A = 5, B = 3, R = 3;
    cout << sum(A, B, R);
 
    return 0;
}

Java

// JAVA implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
 
// Function to return the required sum
static int sum(int A, int B, int R)
{
 
    // To store the sum
    int sum = 0;
 
    // For every row
    for (int i = 1; i <= R; i++) 
    {
 
        // Update the sum as A appears i number
        // of times in the current row
        sum = sum + (i * A);
 
        // Update A for the next row
        A = A + B;
    }
 
    // Return the sum
    return sum;
}
 
// Driver code
public static void main (String[] args) 
              throws java.lang.Exception
{
    int A = 5, B = 3, R = 3;
     
    System.out.print(sum(A, B, R));
}
}
 
// This code is contributed by nidhiva

Python3

# Python3 implementation of the approach
 
# Function to return the required sum
def Sum(A, B, R):
 
    # To store the sum
    ssum = 0
 
    # For every row
    for i in range(1, R + 1):
 
        # Update the sum as A appears i number
        # of times in the current row
        sum = sum + (i * A)
 
        # Update A for the next row
        A = A + B
 
    # Return the sum
    return sum
 
# Driver code
A, B, R = 5, 3, 3
print(Sum(A, B, R))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the required sum
static int sum(int A, int B, int R)
{
 
    // To store the sum
    int sum = 0;
 
    // For every row
    for (int i = 1; i <= R; i++) 
    {
 
        // Update the sum as A appears i number
        // of times in the current row
        sum = sum + (i * A);
 
        // Update A for the next row
        A = A + B;
    }
 
    // Return the sum
    return sum;
}
 
// Driver code
public static void Main () 
{
    int A = 5, B = 3, R = 3;
     
    Console.Write(sum(A, B, R));
}
}
 
// This code is contributed by anuj_67..

Javascript

<script>
// JAVA SCRIPT  implementation of the approach
// Function to return the required sum
function sum( A,  B,  R)
{
 
    // To store the sum
    let sum = 0;
 
    // For every row
    for (let i = 1; i <= R; i++)
    {
 
        // Update the sum as A appears i number
        // of times in the current row
        sum = sum + (i * A);
 
        // Update A for the next row
        A = A + B;
    }
 
    // Return the sum
    return sum;
}
 
// Driver code
 
    let A = 5, B = 3, R = 3;
     
    document.write(sum(A, B, R));
 
//contributed by bobby
 
</script>

Output:

Time Complexity: O(R), since there runs a loop for once from 1 to R.

Auxiliary Space: O(1), since no extra space has been taken.

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