Related Articles
Find sum of product of number in given series
• Last Updated : 23 Oct, 2018

Given two numbers N and T where, and . The task is to find the value of .
Since sum can be large, output it modulo 109+7.

Examples:

Input : 3 2
Output : 38
2*3 + 3*4 + 4*5 = 38

Input : 4 2
Output : 68


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

In the Given Sample Case n = 3 and t = 2.
sum = 2*3+3*4+4*5.
Notice that:    So each term is of the form If we multiply and divide by t! it becomes Which is nothing but Therefore, But we know Therefore So final expression comes out to be But since n is so large we can not calculate it directly, we have to Simplify the above expression.

On Simplifying we get .

Below is the implementation of above approach

## C++

 // C++ program to find sum of product // of number in given series#include using namespace std;  typedef long long ll;const long long MOD = 1000000007;  // function to calculate (a^b)%pll power(ll x, unsigned long long y, ll p){    ll res = 1; // Initialize result      // Update x if it is more than or equal to p    x = x % p;      while (y > 0) {          // If y is odd, multiply x with result        if (y & 1)            res = (res * x) % p;          // y must be even now        y = y >> 1; // y = y/2        x = (x * x) % p;    }      return res;}  // function to return required answerll sumProd(ll n, ll t){    // modulo inverse of denominator    ll dino = power(t + 1, MOD - 2, MOD);      // calculating commentator part    unsigned long long ans = 1;    for (ll i = n + t + 1; i > n; --i)        ans = (ans % MOD * i % MOD) % MOD;      // calculating t!    ll tfact = 1;    for (int i = 1; i <= t; ++i)        tfact = (tfact * i) % MOD;      // accumulating the final answer    ans = ans * dino - tfact + MOD;      return ans % MOD;}int main(){    ll n = 3, t = 2;      // function call to print required sum    cout << sumProd(n, t);      return 0;}

## Java

 // Java program to find sum of product // of number in given series  public class GFG {       static long MOD = 1000000007;           //function to calculate (a^b)%p     static long power(long x, long y, long p)     {      long res = 1; // Initialize result        // Update x if it is more than or equal to p      x = x % p;        while (y > 0) {            // If y is odd, multiply x with result          if ((y & 1)!= 0)              res = (res * x) % p;            // y must be even now          y = y >> 1; // y = y/2          x = (x * x) % p;      }        return res;     }       //function to return required answer     static long sumProd(long n, long t)     {      // modulo inverse of denominator      long dino = power(t + 1, MOD - 2, MOD);        // calculating commentator part      long ans = 1;      for (long i = n + t + 1; i > n; --i)          ans = (ans % MOD * i % MOD) % MOD;        // calculating t!      long tfact = 1;      for (int i = 1; i <= t; ++i)          tfact = (tfact * i) % MOD;        // accumulating the final answer      ans = ans * dino - tfact + MOD;        return ans % MOD;     }       // Driver program    public static void main(String[] args) {                  long n = 3, t = 2;           // function call to print required sum         System.out.println(sumProd(n, t));    }}

## Python3

 # Python 3 program to find sum of product # of number in given series   MOD = 1000000007  # function to calculate (a^b)%pdef power(x, y, p) :      # Initialize result    res = 1      # Update x if it is more than or equal to p    x = x % p      # If y is odd, multiply x with result     while y > 0 :          if y & 1 :            res = (res * x) % p          #  y must be even now        y = y >> 1 # y = y/2        x = (x * x) % p      return res  # function to return required answerdef sumProd(n, t) :      # modulo inverse of denominator     dino = power(t + 1, MOD - 2, MOD)      ans = 1      # calculating commentator part     for i in range(n + t + 1 , n, -1) :        ans = (ans % MOD * i % MOD) % MOD      # calculating t!     tfact = 1    for i in range(1, t+1) :        tfact = (tfact * i) % MOD      # accumulating the final answer     ans = ans * dino - tfact + MOD      return ans % MOD                    # Driver Codeif __name__ == "__main__" :      n, t = 3, 2      # function call to print required sum     print(sumProd(n, t))  # This code is contributed by ANKITRAI1

## C#

 // C# program to find sum of product // of number in given seriesusing System;class GFG {static long MOD = 1000000007;  // function to calculate (a^b)%pstatic long power(long x, long y,                   long p){    long res = 1; // Initialize result          // Update x if it is more    // than or equal to p    x = x % p;          while (y > 0)     {              // If y is odd, multiply x         // with result        if ((y & 1) != 0)            res = (res * x) % p;              // y must be even now        y = y >> 1; // y = y/2        x = (x * x) % p;    }      return res;}  // function to return required answerstatic long sumProd(long n, long t){      // modulo inverse of denominatorlong dino = power(t + 1, MOD - 2, MOD);  // calculating commentator partlong ans = 1;for (long i = n + t + 1; i > n; --i)    ans = (ans % MOD * i % MOD) % MOD;  // calculating t!long tfact = 1;for (int i = 1; i <= t; ++i)    tfact = (tfact * i) % MOD;  // accumulating the final answerans = ans * dino - tfact + MOD;  return ans % MOD;}  // Driver Codepublic static void Main() {    long n = 3, t = 2;      // function call to print required sum    Console.WriteLine(sumProd(n, t));}}  // This code is contributed // by Akanksha Rai(Abby_akku)

## PHP

  0)     {          // If y is odd, multiply         // x with result        if ($y & 1) $res = ($res * $x) % $p;   // y must be even now $y = $y >> 1; // y = y/2 $x = ($x * $x) % $p; }   return $res;}  // function to return required answerfunction sumProd($n, $t){    $MOD = 1000000007;   // modulo inverse of denominator $dino = power($t + 1, $MOD - 2, $MOD);   // calculating commentator part $ans = 1;    for ($i = $n + $t + 1; $i > $n; --$i)        $ans = ($ans % $MOD * $i %                        $MOD) % $MOD;      // calculating t!    $tfact = 1; for ($i = 1; $i <= $t; ++$i) $tfact = ($tfact * $i) % $MOD;   // accumulating the final answer $ans = $ans * $dino - $tfact + $MOD;      return $ans % $MOD;}  // Driver code$n = 3;$t = 2;  // function call to print// required sumecho sumProd($n, $t);  // This code is contributed// by Shivi_Aggarwal?>

Output:

38

Time Complexity: O(T)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up