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Find the sum of elements of the Matrix generated by the given rules
  • Last Updated : 05 Aug, 2019

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:

  1. The first row will contain a single element which is A and the rest of the elements will be 0.
  2. The next row will contain two elements all of which are (A + B) and the rest are 0s.
  3. Third row will contain (A + B + B) three times and the rest are 0s.
  4. …..
  5. 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: Initialise 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++

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// 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;
}

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Java

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// 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

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Python3

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

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C#

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// 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..

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Output:

54

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