Given a positive integer n. The task is to find the sum of even indexed binomial coefficient. That is,
nC0 + nC2 + nC4 + nC6 + nC8 + ………..
Examples :
Input : n = 4
Output : 8
Explanation:
4C0 + 4C2 + 4C4
= 1 + 6 + 1
= 8Input : n = 6
Output : 32
Method 1: (Brute Force)
The idea is to find all the binomial coefficients and find only the sum of even indexed values.
// CPP Program to find sum // of even index term #include <bits/stdc++.h> using namespace std;
// Return the sum of // even index term int evenSum( int n)
{ int C[n + 1][n + 1];
int i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= min(i, n); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using
// previously stored values
else
C[i][j] = C[i - 1][j - 1]
+ C[i - 1][j];
}
}
// finding sum of even index term.
int sum = 0;
for ( int i = 0; i <= n; i += 2)
sum += C[n][i];
return sum;
} // Driver Program int main()
{ int n = 4;
cout << evenSum(n) << endl;
return 0;
} |
// Java Program to find sum // of even index term import java.io.*;
import java.math.*;
class GFG {
// Return the sum of
// even index term
static int evenSum( int n)
{
int C[][] = new int [n + 1 ][n + 1 ];
int i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0 ; i <= n; i++)
{
for (j = 0 ; j <= Math.min(i, n); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1 ;
// else Calculate value using
// previously stored values
else
C[i][j] = C[i - 1 ][j - 1 ]
+ C[i - 1 ][j];
}
}
// finding sum of even index term.
int sum = 0 ;
for (i = 0 ; i <= n; i += 2 )
sum += C[n][i];
return sum;
}
// Driver Program
public static void main(String args[])
{
int n = 4 ;
System.out.println(evenSum(n));
}
} /*This code is contributed by Nikita Tiwari.*/ |
# Python Program to find sum of even index term import math
# Return the sum of even index term def evenSum(n) :
# Creates a list containing n+1 lists,
# each of n+1 items, all set to 0
C = [[ 0 for x in range (n + 1 )] for y in range (n + 1 )]
# Calculate value of Binomial Coefficient
# in bottom up manner
for i in range ( 0 , n + 1 ):
for j in range ( 0 , min (i, n + 1 )):
# Base Cases
if j = = 0 or j = = i:
C[i][j] = 1
# Calculate value using previously
# stored values
else :
C[i][j] = C[i - 1 ][j - 1 ] + C[i - 1 ][j]
# Finding sum of even index term
sum = 0 ;
for i in range ( 0 , n + 1 ):
if n % 2 = = 0 :
sum = sum + C[n][i]
return sum
# Driver method n = 4
print evenSum(n)
# This code is contributed by 'Gitanjali'. |
// C# Program to find sum // of even index term using System;
class GFG {
// Return the sum of
// even index term
static int evenSum( int n)
{
int [,]C = new int [n + 1,n + 1];
int i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.Min(i, n); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i,j] = 1;
// else Calculate value using
// previously stored values
else
C[i,j] = C[i - 1,j - 1]
+ C[i - 1,j];
}
}
// finding sum of even index term.
int sum = 0;
for (i = 0; i <= n; i += 2)
sum += C[n,i];
return sum;
}
// Driver Program
public static void Main()
{
int n = 4;
Console.WriteLine(evenSum(n));
}
} /*This code is contributed by vt_m.*/ |
<?php // PHP Program to find sum // of even index term // Return the sum of // even index term function evenSum( $n )
{ $C = array ( array ());
$i ; $j ;
// Calculate value of Binomial
// Coefficient in bottom up manner
for ( $i = 0; $i <= $n ; $i ++)
{
for ( $j = 0; $j <= min( $i , $n ); $j ++)
{
// Base Cases
if ( $j == 0 or $j == $i )
$C [ $i ][ $j ] = 1;
// Calculate value using
// previously stored values
else
$C [ $i ][ $j ] = $C [ $i - 1][ $j - 1] +
$C [ $i - 1][ $j ];
}
}
// finding sum of even index term.
$sum = 0;
for ( $i = 0; $i <= $n ; $i += 2)
$sum += $C [ $n ][ $i ];
return $sum ;
} // Driver Code $n = 4;
echo evenSum( $n ) ;
// This code is contributed by anuj_67. ?> |
<script> // Javascript Program to find sum // of even index term // Return the sum of // even index term function evenSum(n)
{ var C = Array.from(Array(n+1),
()=> Array(n+1).fill(0));
var i, j;
// Calculate value of Binomial
// Coefficient in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= Math.min(i, n); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using
// previously stored values
else
C[i][j] = C[i - 1][j - 1]
+ C[i - 1][j];
}
}
// finding sum of even index term.
var sum = 0;
for ( var i = 0; i <= n; i += 2)
sum += C[n][i];
return sum;
} // Driver Program var n = 4;
document.write( evenSum(n) ); </script> |
Output
8
Time Complexity: O(n2)
Auxiliary Space: O(n2)
Method 2: (Using Formula)
Sum of even indexed binomial coefficient :
Proof :
We know, (1 + x)n = nC0 + nC1 x + nC2 x2 + ..... + nCn xn Now put x = -x, we get (1 - x)n = nC0 - nC1 x + nC2 x2 + ..... + (-1)nnCn xn Now, adding both the above equation, we get, (1 + x)n + (1 - x)n = 2 * [nC0 + nC2 x2 + nC4 x4 + .......] Put x = 1 (1 + 1)n + (1 - 1)n = 2 * [nC0 + nC2 + nC4 + .......] 2n/2 = nC0 + nC2 + nC4 + ....... 2n-1 = nC0 + nC2 + nC4 + .......
Below is the implementation of this approach :
// CPP Program to find sum even indexed Binomial // Coefficient. #include <bits/stdc++.h> using namespace std;
// Returns value of even indexed Binomial Coefficient // Sum which is 2 raised to power n-1. int evenbinomialCoeffSum( int n)
{ return (1 << (n - 1));
} /* Driver program to test above function*/ int main()
{ int n = 4;
printf ( "%d" , evenbinomialCoeffSum(n));
return 0;
} |
// Java Program to find sum even indexed // Binomial Coefficient. import java.io.*;
class GFG {
// Returns value of even indexed Binomial Coefficient // Sum which is 2 raised to power n-1. static int evenbinomialCoeffSum( int n)
{ return ( 1 << (n - 1 ));
} // Driver Code public static void main(String[] args)
{ int n = 4 ;
System.out.println(evenbinomialCoeffSum(n));
} }
// This code is contributed by 'Gitanjali'. |
# Python program to find sum even indexed # Binomial Coefficient import math
# Returns value of even indexed Binomial Coefficient # Sum which is 2 raised to power n-1. def evenbinomialCoeffSum( n):
return ( 1 << (n - 1 ))
# Driver method if __name__ = = '__main__' :
n = 4
print evenbinomialCoeffSum(n)
# This code is contributed by 'Gitanjali'. |
// C# Program to find sum even indexed // Binomial Coefficient. using System;
class GFG
{ // Returns value of even indexed
// Binomial Coefficient Sum which
// is 2 raised to power n-1.
static int evenbinomialCoeffSum( int n)
{
return (1 << (n - 1));
}
// Driver Code
public static void Main()
{
int n = 4;
Console.WriteLine(evenbinomialCoeffSum(n));
}
} // This code is contributed by 'Vt_m'. |
<?php // PHP Program to find sum // even indexed Binomial // Coefficient. // Returns value of even indexed // Binomial Coefficient Sum which // is 2 raised to power n-1. function evenbinomialCoeffSum( $n )
{ return (1 << ( $n - 1));
} // Driver Code
$n = 4;
echo evenbinomialCoeffSum( $n );
// This code is contributed by anuj_67. ?> |
<script> // JavaScript Program to find sum even indexed // Binomial Coefficient. // Returns value of even indexed Binomial Coefficient // Sum which is 2 raised to power n-1. function evenbinomialCoeffSum(n)
{ return (1 << (n - 1));
} // Driver code let n = 4;
document.write(evenbinomialCoeffSum(n));
// This code is contributed by code_hunt.
</script> |
Output
8
Time Complexity: O(1)
Auxiliary Space: O(1)
Sum of odd index binomial coefficient
Using the above result we can easily prove that the sum of odd index binomial coefficient is also 2n-1.