Sum of product of r and rth Binomial Coefficient (r * nCr)

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Given a positive integer n. The task is to find the sum of the product of r and r^th Binomial Coefficient. In other words find: ? (r * ⁿC_r), where 0 <= r <= n.
Examples:

Input : n = 2
Output : 4
0.²C₀ + 1.²C₁ + 2.²C₂
= 0*2 + 1*2 + 2*1
= 4

Input : n = 5
Output : 80

Method 1 (Brute Force) : The idea is to iterate a loop i from 0 to n and evaluate i * ⁿC_i and add to sum variable.
Below is the implementation of this approach:

C++

// CPP Program to find sum of product of r and
// rth Binomial Coefficient i.e summation r * nCr
#include <bits/stdc++.h>
using namespace std;
#define MAX 100
 
// Return the first n term of binomial coefficient.
void binomialCoeff(int n, int C[])
{
    C[0] = 1; // nC0 is 1
 
    for (int i = 1; i <= n; i++) {
 
        // Compute next row of pascal triangle 
        // using the previous row
        for (int j = min(i, n); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
}
 
// Return summation of r * nCr
int summation(int n)
{
    int C[MAX];
    memset(C, 0, sizeof(C));
 
    // finding the first n term of binomial
    // coefficient
    binomialCoeff(n, C);
 
    // Iterate a loop to find the sum.
    int sum = 0;
    for (int i = 0; i <= n; i++) 
        sum += (i * C[i]);    
 
    return sum;
}
 
// Driven Program
int main()
{
    int n = 2;
    cout << summation(n) << endl;
    return 0;
}

Java

// Java Program to find sum 
// of product of r and rth 
// Binomial Coefficient i.e
// summation r * nCr
class GFG
{
static int MAX = 100;
 
// Return the first n term 
// of binomial coefficient.
static void binomialCoeff(int n, 
                          int C[])
{
    C[0] = 1; // nC0 is 1
 
    for (int i = 1; i <= n; i++)
    {
 
        // Compute next row of
        // pascal triangle using 
        // the previous row
        for (int j = Math.min(i, n); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
}
 
// Return summation
// of r * nCr
static int summation(int n)
{
    int C[] = new int[MAX];
     
    for(int i = 0; i < MAX; i++)
    C[i] = 0;
 
    // finding the first n term  
    // of binomial coefficient
    binomialCoeff(n, C);
 
    // Iterate a loop 
    // to find the sum.
    int sum = 0;
    for (int i = 0; i <= n; i++) 
        sum += (i * C[i]); 
 
    return sum;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 2;
    System.out.println( summation(n));
}
}
 
// This code is contributed by Arnab Kundu

Python 3

# Python 3 Program to find sum of product 
# of r and rth Binomial Coefficient i.e 
# summation r * nCr
MAX = 100
 
# Return the first n term of 
# binomial coefficient.
def binomialCoeff(n, C):
 
    C[0] = 1 # nC0 is 1
 
    for i in range(1, n + 1):
 
        # Compute next row of pascal triangle 
        # using the previous row
        for j in range(min(i, n), -1, -1):
            C[j] = C[j] + C[j - 1]
 
# Return summation of r * nCr
def summation( n):
 
    C = [0] * MAX
 
    # finding the first n term of 
    # binomial coefficient
    binomialCoeff(n, C)
 
    # Iterate a loop to find the sum.
    sum = 0
    for i in range(n + 1): 
        sum += (i * C[i]) 
 
    return sum
 
# Driver Code
if __name__ == "__main__":
     
    n = 2
    print(summation(n))
 
# This code is contributed by ita_c

C#

// C# Program to find sum 
// of product of r and rth 
// Binomial Coefficient i.e
// summation r * nCr
using System;
 
class GFG
{
static int MAX = 100;
 
// Return the first n term 
// of binomial coefficient.
static void binomialCoeff(int n, 
                          int []C)
{
    C[0] = 1; // nC0 is 1
 
    for (int i = 1; i <= n; i++)
    {
 
        // Compute next row of
        // pascal triangle using 
        // the previous row
        for (int j = Math.Min(i, n); 
                 j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
}
 
// Return summation
// of r * nCr
static int summation(int n)
{
    int []C = new int[MAX];
     
    for(int i = 0; i < MAX; i++)
    C[i] = 0;
 
    // finding the first n term 
    // of binomial coefficient
    binomialCoeff(n, C);
 
    // Iterate a loop 
    // to find the sum.
    int sum = 0;
    for (int i = 0; i <= n; i++) 
        sum += (i * C[i]); 
 
    return sum;
}
 
// Driver Code
public static void Main()
{
    int n = 2;
    Console.Write( summation(n));
}
}
 
// This code is contributed 
// by shiv_bhakt

PHP

<?php
// PHP Program to find sum of product 
// of r and rth Binomial Coefficient
// i.e summation r * nCr
$MAX = 100;
 
// Return the first n term of 
// binomial coefficient.
function binomialCoeff($n, &$C)
{
    $C[0] = 1; // nC0 is 1
 
    for ($i = 1; $i <= $n; $i++)
    {
 
        // Compute next row of pascal triangle 
        // using the previous row
        for ($j = min($i, $n); $j > 0; $j--)
            $C[$j] = $C[$j] + $C[$j - 1];
    }
}
 
// Return summation of r * nCr
function summation($n)
{
    global $MAX;
    $C = array_fill(0, $MAX, 0);
 
    // finding the first n term of 
    // binomial coefficient
    binomialCoeff($n, $C);
 
    // Iterate a loop to find the sum.
    $sum = 0;
    for ($i = 0; $i <= $n; $i++) 
        $sum += ($i * $C[$i]); 
 
    return $sum;
}
 
// Driver Code
$n = 2;
echo summation($n);
 
// This code is contributed by mits
?>

Javascript

<script>
 
    // Javascript Program to find sum of product of r and
    // rth Binomial Coefficient i.e summation r * nCr
    MAX = 100
 
    // Return the first n term of binomial coefficient.
    function binomialCoeff(n, C) {
      C[0] = 1; // nC0 is 1
 
      for (var i = 1; i <= n; i++) {
 
        // Compute next row of pascal triangle 
        // using the previous row
        for (var j = Math.min(i, n); j > 0; j--)
          C[j] = C[j] + C[j - 1];
      }
    }
 
    // Return summation of r * nCr
    function summation(n) {
      C = Array(MAX).fill(0);
 
      // finding the first n term of binomial
      // coefficient
      binomialCoeff(n, C);
 
      // Iterate a loop to find the sum.
      var sum = 0;
      for (var i = 0; i <= n; i++)
        sum += (i * C[i]);
 
      return sum;
    }
 
    // Driven Program
    var n = 2;
    document.write(summation(n));
     
    // This code is contributed by noob2000.
  </script>

Output:

Time complexity: O(n*n)

space complexity: O(MAX)

Method 2 (Using formula) :
Mathematically we need to find,
? (i * ⁿC_i), where 0 <= i <= n
= ? (ⁱC₁ * ⁿC_i), (Since ⁿC₁ = n, we can write i as ⁱC₁)
= ? ( (i! / (i – 1)! * 1!) * (n! / (n – i)! * i!)
On cancelling i! from numerator and denominator
= ? (n! / (i – 1)! * (n – i)!)
= ? n * ((n – 1)!/ (i – 1)! * (n – i)!)
(Using reverse of ⁿC_r = (n)!/ (r)! * (n – r)!)
= n * ? ^{n – 1}C_{r – 1}
= n * 2^{n – 1} (Since ? ⁿC_r = 2ⁿ)
Below is the implementation of this approach:

C++

// CPP Program to find sum of product of r and
// rth Binomial Coefficient i.e summation r * nCr
#include <bits/stdc++.h>
using namespace std;
#define MAX 100
 
// Return summation of r * nCr
int summation(int n)
{
    return n << (n - 1);
}
 
// Driven Program
int main()
{
    int n = 2;
    cout << summation(n) << endl;
    return 0;
}

Java

// Java Program to find sum of product of
// r and rth Binomial Coefficient i.e 
// summation r * nCr
import java.io.*;
 
class GFG {
 
    static int MAX = 100;
     
    // Return summation of r * nCr
    static int summation(int n)
    {
        return n << (n - 1);
    }
     
    // Driven Program
    public static void main (String[] args)
    {
        int n = 2;
        System.out.println( summation(n));
    }
}
 
// This code is contributed by anuj_67.

Python3

# Python3 Program to 
# find sum of product 
# of r and rth Binomial 
# Coefficient i.e 
# summation r * nCr
 
# Return summation 
# of r * nCr
def summation( n):
    return n << (n - 1);
 
# Driver Code
n = 2;
print(summation(n));
 
# This code is contributed 
# by mits

C#

// C# Program to find sum of product of
// r and rth Binomial Coefficient i.e 
// summation r * nCr
using System;
 
class GFG {
 
    //static int MAX = 100;
     
    // Return summation of r * nCr
    static int summation(int n)
    {
        return n << (n - 1);
    }
     
    // Driver Code
    public static void Main ()
    {
        int n = 2;
        Console.WriteLine( summation(n));
    }
}
 
// This code is contributed by anuj_67.

PHP

<?php
// PHP Program to find sum 
// of product of r and
// rth Binomial Coefficient
// i.e summation r * nCr
 
// Return summation of r * nCr
function summation( $n)
{
    return $n << ($n - 1);
}
 
// Driver Code
$n = 2;
echo summation($n) ;
 
// This code is contributed 
// by shiv_bhakt
?>

Javascript

<script>
 
// Javascript Program to find sum of product of r and 
// rth Binomial Coefficient i.e summation r * nCr 
var MAX=100 
 
// Return summation of r * nCr 
function summation(n) 
{ 
    return n << (n - 1); 
} 
 
// Driven Program 
var n = 2; 
document.write(summation(n));
     
</script>

Output:

Time Complexity: O(1)

Auxiliary Space: O(1)

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