# Sum of product of consecutive Binomial Coefficients

• Difficulty Level : Medium
• Last Updated : 12 Jul, 2022

Given a positive integer n. The task is to find the sum of product of consecutive binomial coefficient i.e
nC0*nC1 + nC1*nC2 + ….. + nCn-1*nCn

Examples:

```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```

Method 1: The idea is to find all the binomial coefficients up to nth term and find the sum of the product of consecutive coefficients.

Below is the implementation of this approach:

## 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 = 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 = 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``?>`

## Javascript

 ``

Output

`15`

Method 2:
We know,
(1 + x)n = nC0 + nC1*x + nC2*x2 + …. + nCn*xn … (1)
(1 + 1/x)n = nC0 + nC1/x + nC2/x2 + …. + nCn/xn … (2)
Multiplying (1) and (2), we get
(1 + x)2n/xn = (nC0 + nC1*x + nC2*x2 + …. + nCn*xn) * (nC0 + nC1/x + nC2/x2 + …. + nCn/xn)
(2nC0 + 2nC1*x + 2nC2*x2 + …. + 2nCn*xn)/xn = (nC0 + nC1*x + nC2*x2 + …. + nCn*xn) * (nC0 + nC1/x + nC2/x2 + …. + nCn/xn)
Now, find the coefficient of x in LHS,
Observe rth term of expansion in numerator is 2nCrxr
To find the coefficient of x in (1 + x)2n/xn, r should be n + 1, because power of x in denominator will reduce it.
So, coefficient of x in LHS = 2nCn + 1 or 2nCn – 1
Now, find the coefficient of x in RHS,
r th term of first expansion of multiplication is nCr * xr
t th term of second expansion of multiplication is nCt / xt
So term after multiply will be nCr * xr * nCt / xt or
nCr * nCt * xr / xt
Put r = t + 1, we get,
nCt+1 * nCt * x
Observe there will be n such term in the expansion of multiply, so t range from 0 to n – 1.
Therefore, coefficient of x in RHS = nC0*nC1 + nC1*nC2 + ….. + nCn-1*nCn
Comparing coefficient of x in LHS and RHS, we can say,
nC0*nC1 + nC1*nC2 + ….. + nCn-1*nCn = 2nCn – 1

Below is implementation of this approach:

## 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 = 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 = 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.`

## 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``?>`

## Javascript

 ``

Output:

`15`

Time Complexity: O(n*k)

Auxiliary Space: O(k)

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