Program to find sum of the given sequence

Given two numbers $N$ and $K$ . The task is to find the sum of the sequence given below.

(1*2*3*…*k) + (2*3*…*k*(k+1)) + (3*4*..*(k+1)*(k+2)) +…..+((n-k+1)*(n-k+2)*…*(n-k+k)).

Since the output can be large, print the answer under modulo 10^9+7.

Examples:

Input : N = 3, K = 2
Output : 8

Input : N = 4, K = 2
Output : 20

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Let us take the given example and try to reduce it to a general formula.

In the given example for n = 3 and k=2,

Sum = 1*2 + 2*3

We know that:
$$1\times2=2!\times 0!$\\ $2\times3=3!\times1!$\\$

So each term is of the form:

$\frac{x!}{\left(x-k\right)!}$

If we multiply and divide by $k!$ , it becomes

$k!\cdot\frac{x!}{\left(x-k\right)!\cdot k!}$

Which is nothing but,
$$k!\times$ ${N}\choose{k}$\\$

Therefore,
$sum = $k\times \sum_{x=k}^{n}$ $ {x}\choose{k}$ =$ {n+1}\choose{k+1}$ $\times k! $\\$

But since n is so large we can not calculate it directly, we have to simplify the above expression.

On Simplifying we get,

$\frac{\left(n\right) \cdot \left(n-1\right) \cdot \left(n-2\right) \dots \left(n-k+1\right)}{k+1}$

Below is the implementation of the above idea:

C++

// CPP program to find the sum of the 
// given sequence 
  
#include <bits/stdc++.h> 
using namespace std; 
  
const long long MOD = 1000000007; 
  
// function to find moudulo inverse 
// under 10^9+7 
long long modInv(long long x) 
{ 
    long long n = MOD - 2; 
    long long result = 1; 
    while (n) { 
        if (n & 1) 
            result = result * x % MOD; 
        x = x * x % MOD; 
        n = n / 2; 
    } 
      
    return result; 
} 
  
// Function to find the sum of the  
// given sequence 
long long getSum(long long n, long long k) 
{ 
    long long ans = 1; 
      
    for (long long i = n + 1; i > n - k; i--) 
        ans = ans * i % MOD; 
    ans = ans * modInv(k + 1) % MOD; 
      
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    long long n = 3, k = 2; 
      
    cout<<getSum(n,k); 
      
    return 0; 
} 

Java

// Java program to find the sum of the  
// given sequence  
  
class GFG { 
  
    static long MOD = 1000000007; 
  
// function to find moudulo inverse  
// under 10^9+7  
    static long modInv(long x) { 
        long n = MOD - 2; 
        long result = 1; 
        while (n > 0) { 
            if ((n & 1) > 0) { 
                result = result * x % MOD; 
            } 
            x = x * x % MOD; 
            n = n / 2; 
        } 
  
        return result; 
    } 
  
// Function to find the sum of the  
// given sequence  
    static long getSum(long n, long k) { 
        long ans = 1; 
  
        for (long i = n + 1; i > n - k; i--) { 
            ans = ans * i % MOD; 
        } 
        ans = ans * modInv(k + 1) % MOD; 
  
        return ans; 
    } 
  
// Driver code  
    public static void main(String[] args) { 
        long n = 3, k = 2; 
        System.out.println(getSum(n, k)); 
    } 
} 

Python3

# Python3 program to find the sum  
# of the given sequence 
  
MOD = 1000000007; 
  
# function to find moudulo inverse 
# under 10^9+7 
def modInv(x): 
  
    n = MOD - 2; 
    result = 1; 
    while (n): 
        if (n&1): 
            result = result * x % MOD; 
        x = x * x % MOD; 
        n = int(n / 2); 
      
    return result; 
  
# Function to find the sum of  
# the given sequence 
def getSum(n, k): 
  
    ans = 1; 
      
    for i in range(n + 1, n - k, -1): 
        ans = ans * i % MOD; 
    ans = ans * modInv(k + 1) % MOD; 
      
    return ans; 
  
# Driver code 
n = 3; 
k = 2; 
      
print(getSum(n,k)); 
      
# This code is contributed by mits 

C#

// C# program to find the sum of the 
// given sequence 
using System; 
  
   
  
// function to find moudulo inverse 
// under 10^9+7 
class gfg 
{ 
    public long MOD = 1000000007; 
 public long modInv(long x) 
 { 
    long n = MOD - 2; 
    long result = 1; 
    while (n >0) { 
        if ((n & 1) > 0) 
            result = result * x % MOD; 
        x = x * x % MOD; 
        n = n / 2; 
    } 
      
    return result; 
} 
  
// Function to find the sum of the  
// given sequence 
  public long getSum(long n, long k) 
  { 
    long ans = 1; 
      
    for (long i = n + 1; i > n - k; i--) 
        ans = ans * i % MOD; 
    ans = ans * modInv(k + 1) % MOD; 
      
    return ans; 
 } 
} 
  
// Driver code 
class geek 
{ 
  public static int Main() 
 { 
     gfg g = new gfg(); 
    long n = 3, k = 2; 
      
    Console.WriteLine(g.getSum(n,k)); 
      
    return 0; 
 } 
} 
//This code is contributed by SoumikMondal 

PHP

<?php 
// PHP program to find the sum of  
// the given sequence 
  
// function to find moudulo inverse 
// under 10^9+7 
function modInv($x) 
{ 
    $MOD = 1000000007; 
    $n = $MOD - 2; 
    $result = 1; 
    while ($n)  
    { 
        if ($n & 1) 
            $result = $result * $x % $MOD; 
        $x = $x * $x % $MOD; 
        $n = $n / 2; 
    } 
      
    return $result; 
} 
  
// Function to find the sum of the  
// given sequence 
function getSum($n, $k) 
{ 
    $MOD = 1000000007; 
    $ans = 1; 
      
    for ($i = $n + 1; $i > $n - $k; $i--) 
        $ans = $ans * $i % $MOD; 
    $ans = $ans * modInv($k + 1) % $MOD; 
      
    return $ans; 
} 
  
// Driver code 
$n = 3; $k = 2; 
      
echo getSum($n, $k); 
  
// This code is contributed  
// by Akanksha Rai 
?> 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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