给定一个正整数 N .任务是求偶数指数二项系数之和。就是， N C 0 + N C 2. + N C 4. + N C 6. + N C 8. + ……….. 例如：
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Input : n = 4Output : 84C0 + 4C2 + 4C4= 1 + 6 + 1= 8Input : n = 6Output : 32
方法1：（暴力） 其思想是找到所有的二项式系数，只找到偶数索引值之和。
CPP
// 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
// 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
# 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#
// 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 // 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. ?> |
Javascript
<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> |
输出：
8
时间复杂性： O（n） 2. ) 方法2：（使用公式） 偶数指数二项系数之和： 
证据：
We know,(1 + x)n = nC0 + nC1 x + nC2 x2 + ..... + nCn xnNow put x = -x, we get(1 - x)n = nC0 - nC1 x + nC2 x2 + ..... + (-1)n nCn xnNow, 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 + .......
以下是该方法的实施情况：
C++
// 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
// 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
# 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#
// 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 // 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. ?> |
Javascript
<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> |
输出：
8
时间复杂性： O（1） 奇指数二项系数之和 利用上述结果，我们可以很容易地证明奇指数二项系数之和也是2 n-1 .
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