连续二项式系数的乘积和

给定正整数n 。任务是找到连续二项式系数的乘积之和，即
ⁿ C ₀ * ⁿ C ₁ + ⁿ C ₁ * ⁿ C ₂ +….. + ⁿ C _n-1 * ⁿ C _n

例子：

Input : n = 3
Output : 15
3C0*3C1 + 3C1*3C2 +3C2*3C3
= 1*3 + 3*3 + 3*1
= 3 + 9 + 3
= 15

Input : n = 4
Output : 56

方法1：想法是找到直到第n个项的所有二项式系数，并找到连续系数乘积的和。

以下是此方法的实现：

C++

// CPP Program to find sum of product of
// consecutive Binomial Coefficient.
#include 
using namespace std;
#define MAX 100
  
// Find the binomial coefficient upto nth term
void binomialCoeff(int C[], int n)
{
    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 the sum of the product of
// consecutive binomial coefficient.
int sumOfproduct(int n)
{
    int sum = 0;
    int C[MAX] = { 0 };
  
    binomialCoeff(C, n);
  
    // finding the sum of product of 
    // consecutive coefficient.
    for (int i = 0; i <= n; i++) 
        sum += C[i] * C[i + 1];    
  
    return sum;
}
  
// Driven Program
int main()
{
    int n = 3;
    cout << sumOfproduct(n) << endl;
    return 0;
}

Java

// Java Program to find sum of product of
// consecutive Binomial Coefficient.
  
import java.io.*;
  
class GFG {
     
static int  MAX = 100;
  
// Find the binomial coefficient upto nth term
static void binomialCoeff(int C[], int n)
{
    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 the sum of the product of
// consecutive binomial coefficient.
static int sumOfproduct(int n)
{
    int sum = 0;
    int C[] = new int[MAX];
  
    binomialCoeff(C, n);
  
    // finding the sum of product of 
    // consecutive coefficient.
    for (int i = 0; i <= n; i++) 
        sum += C[i] * C[i + 1]; 
  
    return sum;
}
  
// Driven Program
  
    public static void main (String[] args) {
    int n = 3;
    System.out.println( sumOfproduct(n));
    }
}
   
// This code is contributed by inder_verma..

Python3

# Python3 Program to find sum of product 
# of consecutive Binomial Coefficient.
MAX = 100;
  
# Find the binomial coefficient upto 
# nth term
def binomialCoeff(C, n):
  
    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), 0, -1):
            C[j] = C[j] + C[j - 1];
      
    return C;
  
# Return the sum of the product of 
# consecutive binomial coefficient.
def sumOfproduct(n):
  
    sum = 0;
    C = [0] * MAX;
  
    C = binomialCoeff(C, n);
  
    # finding the sum of 
    # product of consecutive
    # coefficient.
    for i in range(n + 1): 
        sum += C[i] * C[i + 1]; 
  
    return sum;
  
# Driver Code
n = 3;
print(sumOfproduct(n));
  
# This code is contributed by mits

C#

// C# Program to find sum of 
// product of consecutive
// Binomial Coefficient.
using System;
  
class GFG 
{
static int MAX = 100;
  
// Find the binomial coefficient
// upto nth term
static void binomialCoeff(int []C, int n)
{
    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 the sum of the product of
// consecutive binomial coefficient.
static int sumOfproduct(int n)
{
    int sum = 0;
    int []C = new int[MAX];
  
    binomialCoeff(C, n);
  
    // finding the sum of product of 
    // consecutive coefficient.
    for (int i = 0; i <= n; i++) 
        sum += C[i] * C[i + 1]; 
  
    return sum;
}
  
// Driven Code
public static void Main ()
{
    int n = 3;
    Console.WriteLine(sumOfproduct(n));
}
}
  
// This code is contributed by anuj_67

PHP

 0; $j--)
            $C[$j] = $C[$j] +
                     $C[$j - 1];
    }
    return $C;
}
  
// Return the sum of the 
// product of consecutive 
// binomial coefficient.
function sumOfproduct($n)
{
    global $MAX;
    $sum = 0;
    $C = array_fill(0, $MAX, 0);
  
    $C = binomialCoeff($C, $n);
  
    // finding the sum of 
    // product of consecutive
    // coefficient.
    for ($i = 0; $i <= $n; $i++) 
        $sum += $C[$i] * $C[$i + 1]; 
  
    return $sum;
}
  
// Driver Code
$n = 3;
echo sumOfproduct($n);
  
// This code is contributed by mits
?>

输出

15

方法2：
我们知道，
(1 + x) ⁿ = ⁿ C ₀ + ⁿ C ₁ * x + ⁿ C ₂ * x ² +…。 + ⁿ C _n * x ⁿ …(1)

(1 + 1 / x) ⁿ = ⁿ C ₀ + ⁿ C ₁ / x + ⁿ C ₂ / x ² +…。 + ⁿ C _n / x ⁿ …(2)

分别乘以(1)和(2)
(1 + x) ²ⁿ / x ⁿ =( ⁿ C ₀ + ⁿ C ₁ * x + ⁿ C ₂ * x ² +…。+ ⁿ C _n * x ⁿ )*( ⁿ C ₀ + ⁿ C ₁ / x + ⁿ C ₂ / x ² +…。+ ⁿ C _n / x ⁿ )

( ²ⁿ C ₀ + ²ⁿ C ₁ * x + ²ⁿ C ₂ * x ² +…。+ ²ⁿ C _n * x ⁿ )/ x ⁿ =( ⁿ C ₀ + ⁿ C ₁ * x + ⁿ C ₂ * x ² + …。+ ⁿ C _n * x ⁿ )*( ⁿ C ₀ + ⁿ C ₁ / x + ⁿ C ₂ / x ² +…。+ ⁿ C _n / x ⁿ )

现在，在LHS中找到x的系数，
观察分子的第r个扩展项是²ⁿ C _r x ^r 。
为了在(1 + x) ²ⁿ / x ^n中找到x的系数，r应该是n + 1，因为x在分母中的幂会减小它。
因此，在LHS中x的系数= ²ⁿ C _{n +1}或²ⁿ C _{n – 1}

现在，在RHS中找到x的系数，
乘法的第一个展开的第r个项是ⁿ C _r * x ^r
乘法第二次扩展的第t个项是ⁿ C _t / x ^t
因此乘法后的项将是ⁿ C _r * x ^r * ⁿ C _t / x ^t或
ⁿ C _r * ⁿ C _t * x ^r / x ^t
令r = t + 1，我们得到，
ⁿ C _{t + 1} * ⁿ C _t * x
观察到乘法的扩展中将有n个这样的项，因此t的范围是0到n – 1。
因此，RHS中的x系数= ⁿ C ₀ * ⁿ C ₁ + ⁿ C ₁ * ⁿ C ₂ +….. + ⁿ C _n-1 * ⁿ C _n

比较LHS和RHS中x的系数，可以说，
ⁿ C ₀ * ⁿ C ₁ + ⁿ C ₁ * ⁿ C ₂ +….. + ⁿ C _n-1 * ⁿ C _n = ²ⁿ C _{n – 1}

以下是此方法的实现：

C++

// CPP Program to find sum of product of 
// consecutive Binomial Coefficient.
#include 
using namespace std;
#define MAX 100
  
// Find the binomial coefficient up to nth 
// term
int binomialCoeff(int n, int k)
{
    int C[k + 1];
    memset(C, 0, sizeof(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, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}
  
// Return the sum of the product of
// consecutive binomial coefficient.
int sumOfproduct(int n)
{
    return binomialCoeff(2 * n, n - 1);
}
  
// Driven Program
int main()
{
    int n = 3;
  
    cout << sumOfproduct(n) << endl;
    return 0;
}

Java

// Java Program to find sum of 
// product of consecutive 
// Binomial Coefficient.
import java.io.*;
  
class GFG 
{
    static int MAX = 100;
      
    // Find the binomial coefficient 
    // up to nth term
    static int binomialCoeff(int n, 
                             int k)
    {
        int C[] = new int[k + 1];
          
        // memset(C, 0, sizeof(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, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
    }
      
    return C[k];
}
  
// Return the sum of the 
// product of consecutive
// binomial coefficient.
static int sumOfproduct(int n)
{
    return binomialCoeff(2 * n, 
                         n - 1);
}
  
// Driver Code
public static void main (String[] args) 
{
    int n = 3;
    System.out.println(sumOfproduct(n));
}
}
  
// This code is contributed
// by shiv_bhakt.

Python3

# Python3 Program to find sum of product 
# of consecutive Binomial Coefficient.
MAX = 100;
  
# Find the binomial coefficient 
# up to nth term
def binomialCoeff(n, k):
  
    C = [0] * (k + 1);
  
    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, k), 0, -1):
            C[j] = C[j] + C[j - 1];
    return C[k];
  
# Return the sum of the product of
# consecutive binomial coefficient.
def sumOfproduct(n):
    return binomialCoeff(2 * n, n - 1);
  
# Driver Code
n = 3;
print(sumOfproduct(n));
  
# This code is contributed by mits

C＃

// C# Program to find sum of 
// product of consecutive 
// Binomial Coefficient.
using System;
  
class GFG
{
      
    // Find the binomial 
    // coefficient up to 
    // nth term
    static int binomialCoeff(int n, 
                             int k)
    {
        int []C = new int[k + 1];
          
        // memset(C, 0, sizeof(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, k); 
                             j > 0; j--)
                C[j] = C[j] + C[j - 1];
    }
      
    return C[k];
}
  
// Return the sum of the 
// product of consecutive
// binomial coefficient.
static int sumOfproduct(int n)
{
    return binomialCoeff(2 * n, 
                         n - 1);
}
  
// Driver Code
static public void Main ()
{
    int n = 3;
    Console.WriteLine(sumOfproduct(n));
}
}
  
// This code is contributed
// by @ajit.

的PHP

 0; $j--)
            $C[$j] = $C[$j] + $C[$j - 1];
    }
    return $C[$k];
}
  
// Return the sum of the product of
// consecutive binomial coefficient.
function sumOfproduct($n)
{
    return binomialCoeff(2 * $n, $n - 1);
}
  
// Driver Code
$n = 3;
echo sumOfproduct($n);
  
// This code is contributed by mits
?>

输出：

15