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
= 8
Input : 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
#include <bits/stdc++.h>
using namespace std;
int evenSum( int n)
{
int C[n + 1][n + 1];
int i, j;
for (i = 0; i <= n; i++) {
for (j = 0; j <= min(i, n); j++) {
if (j == 0 || j == i)
C[i][j] = 1;
else
C[i][j] = C[i - 1][j - 1]
+ C[i - 1][j];
}
}
int sum = 0;
for ( int i = 0; i <= n; i += 2)
sum += C[n][i];
return sum;
}
int main()
{
int n = 4;
cout << evenSum(n) << endl;
return 0;
}
|
Java
import java.io.*;
import java.math.*;
class GFG {
static int evenSum( int n)
{
int C[][] = new int [n + 1 ][n + 1 ];
int i, j;
for (i = 0 ; i <= n; i++)
{
for (j = 0 ; j <= Math.min(i, n); j++)
{
if (j == 0 || j == i)
C[i][j] = 1 ;
else
C[i][j] = C[i - 1 ][j - 1 ]
+ C[i - 1 ][j];
}
}
int sum = 0 ;
for (i = 0 ; i <= n; i += 2 )
sum += C[n][i];
return sum;
}
public static void main(String args[])
{
int n = 4 ;
System.out.println(evenSum(n));
}
}
|
Python
import math
def evenSum(n) :
C = [[ 0 for x in range (n + 1 )] for y in range (n + 1 )]
for i in range ( 0 , n + 1 ):
for j in range ( 0 , min (i, n + 1 )):
if j = = 0 or j = = i:
C[i][j] = 1
else :
C[i][j] = C[i - 1 ][j - 1 ] + C[i - 1 ][j]
sum = 0 ;
for i in range ( 0 , n + 1 ):
if n % 2 = = 0 :
sum = sum + C[n][i]
return sum
n = 4
print evenSum(n)
|
C#
using System;
class GFG {
static int evenSum( int n)
{
int [,]C = new int [n + 1,n + 1];
int i, j;
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.Min(i, n); j++)
{
if (j == 0 || j == i)
C[i,j] = 1;
else
C[i,j] = C[i - 1,j - 1]
+ C[i - 1,j];
}
}
int sum = 0;
for (i = 0; i <= n; i += 2)
sum += C[n,i];
return sum;
}
public static void Main()
{
int n = 4;
Console.WriteLine(evenSum(n));
}
}
|
PHP
<?php
function evenSum( $n )
{
$C = array ( array ());
$i ; $j ;
for ( $i = 0; $i <= $n ; $i ++)
{
for ( $j = 0; $j <= min( $i , $n ); $j ++)
{
if ( $j == 0 or $j == $i )
$C [ $i ][ $j ] = 1;
else
$C [ $i ][ $j ] = $C [ $i - 1][ $j - 1] +
$C [ $i - 1][ $j ];
}
}
$sum = 0;
for ( $i = 0; $i <= $n ; $i += 2)
$sum += $C [ $n ][ $i ];
return $sum ;
}
$n = 4;
echo evenSum( $n ) ;
?>
|
Javascript
<script>
function evenSum(n)
{
var C = Array.from(Array(n+1),
()=> Array(n+1).fill(0));
var i, j;
for (i = 0; i <= n; i++) {
for (j = 0; j <= Math.min(i, n); j++) {
if (j == 0 || j == i)
C[i][j] = 1;
else
C[i][j] = C[i - 1][j - 1]
+ C[i - 1][j];
}
}
var sum = 0;
for ( var i = 0; i <= n; i += 2)
sum += C[n][i];
return sum;
}
var n = 4;
document.write( evenSum(n) );
</script>
|
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)n nCn 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 :
C++
#include <bits/stdc++.h>
using namespace std;
int evenbinomialCoeffSum( int n)
{
return (1 << (n - 1));
}
int main()
{
int n = 4;
printf ( "%d" , evenbinomialCoeffSum(n));
return 0;
}
|
Java
import java.io.*;
class GFG {
static int evenbinomialCoeffSum( int n)
{
return ( 1 << (n - 1 ));
}
public static void main(String[] args)
{
int n = 4 ;
System.out.println(evenbinomialCoeffSum(n));
}
}
|
Python
import math
def evenbinomialCoeffSum( n):
return ( 1 << (n - 1 ))
if __name__ = = '__main__' :
n = 4
print evenbinomialCoeffSum(n)
|
C#
using System;
class GFG
{
static int evenbinomialCoeffSum( int n)
{
return (1 << (n - 1));
}
public static void Main()
{
int n = 4;
Console.WriteLine(evenbinomialCoeffSum(n));
}
}
|
PHP
<?php
function evenbinomialCoeffSum( $n )
{
return (1 << ( $n - 1));
}
$n = 4;
echo evenbinomialCoeffSum( $n );
?>
|
Javascript
<script>
function evenbinomialCoeffSum(n)
{
return (1 << (n - 1));
}
let n = 4;
document.write(evenbinomialCoeffSum(n));
</script>
|
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.