Given a natural number N, the task is to find the sum of the series up to Nth term where the ith term denotes the difference between the product of first i natural numbers and sum of first i natural numbers, i.e.,
{ 1 – 1 } + { (1 * 2) – (2 + 1) } + { (1 * 2 * 3) – (3 + 2 + 1) } + { (1 * 2 * 3 * 4) – (4 + 3 + 2 + 1) } + ………
Examples:
Input: 2
Output: -1
Explanation: { 1 – 1 } + { (1 * 2) – (2 + 1) } = { 0 } + { -1 } = -1Input: 4
Output: 13
Explanation:
{ 0 } + { (1 * 2) – (2 + 1) } + { (1 * 2 * 3) – (3 + 2 + 1) } + { (1 * 2 * 3 * 4) – (4 + 3 + 2 + 1) }
= { 0 } + { -1 } + { 0 } + { 14 }
= 0 – 1 + 0 + 14
= 13
Iterative Approach:
Follow the steps below to solve the problem:
- Initialize three variables:
- currProd: Stores the product of all natural numbers upto current term.
- currSum: Stores the sum of all natural numbers upto current term.
- sum: Stores the sum of the series upto current term
- Initialize currSum = 1, currSum = 1, and sum = 0 (Since, 1 – 1 = 0) to store their respective values for the first term.
- Iterate over the range [2, N] and update the following for every ith iteration:
- currSum = currSum + i
- currProd = currProd * i
- sum = currProd – currSum
- Return the final value of sum.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to// calculate the// sum upto Nth termint seriesSum(int n){ // Stores the sum // of the series int sum = 0; // Stores the // product of // natural numbers // upto the // current term int currProd = 1; // Stores the sum // of natural numbers // upto the // upto current term int currSum = 1; // Generate the remaining // terms and calculate sum for (int i = 2; i <= n; i++) { currProd *= i; currSum += i; // Update the sum sum += currProd - currSum; } // Return the sum return sum;}// Driver Programint main(){ int N = 5; cout << seriesSum(N) << " ";} |
Java
// Java program to implement the // above approachimport java.lang.*;class GFG{// Function to calculate the// sum upto Nth termstatic int seriesSum(int n){ // Stores the sum // of the series int sum = 0; // Stores the product of // natural numbers upto // the current term int currProd = 1; // Stores the sum of natural // numbers upto the upto // current term int currSum = 1; // Generate the remaining // terms and calculate sum for(int i = 2; i <= n; i++) { currProd *= i; currSum += i; // Update the sum sum += currProd - currSum; } // Return the sum return sum;}// Driver codepublic static void main(String[] args){ int N = 5; System.out.print(seriesSum(N));}}// This code is contributed by rock_cool |
Python3
# Python3 program to implement# the above approach# Function to calculate the# sum upto Nth termdef seriesSum(n): # Stores the sum # of the series sum1 = 0; # Stores the product of # natural numbers upto the # current term currProd = 1; # Stores the sum of natural numbers # upto the upto current term currSum = 1; # Generate the remaining # terms and calculate sum for i in range(2, n + 1): currProd *= i; currSum += i; # Update the sum sum1 += currProd - currSum; # Return the sum return sum1;# Driver CodeN = 5;print(seriesSum(N), end = " ");# This code is contributed by Code_Mech |
C#
// C# program to implement the // above approachusing System;class GFG{// Function to calculate the// sum upto Nth termstatic int seriesSum(int n){ // Stores the sum // of the series int sum = 0; // Stores the product of // natural numbers upto // the current term int currProd = 1; // Stores the sum of natural // numbers upto the upto // current term int currSum = 1; // Generate the remaining // terms and calculate sum for(int i = 2; i <= n; i++) { currProd *= i; currSum += i; // Update the sum sum += currProd - currSum; } // Return the sum return sum;}// Driver codepublic static void Main(){ int N = 5; Console.Write(seriesSum(N));}}// This code is contributed by Code_Mech |
Javascript
<script> // Javascript program to implement // the above approach // Function to // calculate the // sum upto Nth term function seriesSum(n) { // Stores the sum // of the series let sum = 0; // Stores the // product of // natural numbers // upto the // current term let currProd = 1; // Stores the sum // of natural numbers // upto the // upto current term let currSum = 1; // Generate the remaining // terms and calculate sum for (let i = 2; i <= n; i++) { currProd *= i; currSum += i; // Update the sum sum += currProd - currSum; } // Return the sum return sum; } let N = 5; document.write(seriesSum(N) + " "); </script> |
Output:
118
Time Complexity: O(N)
Auxiliary Space: O(1)
Recursive Approach:
- Recursively calculate the sum by calling the function updating K (denotes the current term).
- Update precomputed prevSum by adding multi * K – add + K
- Recursively compute for all values of K updating prevSum, multi and add at every iteration.
- Finally, return the value of prevSum.
Below is the implementation of the above approach:
C++
// C++ program to calculate the// sum upto the Nth term of the// given series#include <bits/stdc++.h>using namespace std;// Recursive Function to calculate the// sum upto Nth termint seriesSumUtil(int k, int n, int prevSum, int multi, int add){ // If N-th term is calculated if (k == n + 1) { return prevSum; } // Update multi to store // product upto K multi = multi * k; // Update add to store // sum upto K add = add + k; // Update prevSum to store sum // upto K prevSum = prevSum + multi - add; // Proceed to next K return seriesSumUtil(k + 1, n, prevSum, multi, add);}// Function to calculate// and return the// Sum upto Nth termint seriesSum(int n){ if (n == 1) return 0; int prevSum = 0; int multi = 1; int add = 1; // Recursive Function return seriesSumUtil(2, n, prevSum, multi, add);}// Driver Programint main(){ int N = 5; cout << seriesSum(N) << " ";} |
Java
// Java program to calculate the// sum upto the Nth term of the// given seriesclass GFG{ // Recursive Function to calculate the// sum upto Nth termstatic int seriesSumUtil(int k, int n, int prevSum, int multi, int add){ // If N-th term is calculated if (k == n + 1) { return prevSum; } // Update multi to store // product upto K multi = multi * k; // Update add to store // sum upto K add = add + k; // Update prevSum to store sum // upto K prevSum = prevSum + multi - add; // Proceed to next K return seriesSumUtil(k + 1, n, prevSum, multi, add);}// Function to calculate and // return the Sum upto Nth termstatic int seriesSum(int n){ if (n == 1) return 0; int prevSum = 0; int multi = 1; int add = 1; // Recursive Function return seriesSumUtil(2, n, prevSum, multi, add);}// Driver codepublic static void main(String []args){ int N = 5; System.out.println(seriesSum(N)); }}// This code is contributed by Ritik Bansal |
Python3
# Python3 program to calculate the# sum upto the Nth term of the# given series# Recursive Function to calculate the# sum upto Nth termdef seriesSumUtil(k, n, prevSum, multi, add): # If N-th term is calculated if (k == n + 1): return prevSum; # Update multi to store # product upto K multi = multi * k; # Update add to store # sum upto K add = add + k; # Update prevSum to store sum # upto K prevSum = prevSum + multi - add; # Proceed to next K return seriesSumUtil(k + 1, n, prevSum, multi, add);# Function to calculate and# return the Sum upto Nth termdef seriesSum(n): if (n == 1): return 0; prevSum = 0; multi = 1; add = 1; # Recursive Function return seriesSumUtil(2, n, prevSum, multi, add);# Driver codeif __name__ == '__main__': N = 5; print(seriesSum(N));# This code is contributed by Princi Singh |
C#
// C# program to calculate the// sum upto the Nth term of the// given seriesusing System;class GFG{ // Recursive Function to calculate the// sum upto Nth termstatic int seriesSumUtil(int k, int n, int prevSum, int multi, int add){ // If N-th term is calculated if (k == n + 1) { return prevSum; } // Update multi to store // product upto K multi = multi * k; // Update add to store // sum upto K add = add + k; // Update prevSum to store sum // upto K prevSum = prevSum + multi - add; // Proceed to next K return seriesSumUtil(k + 1, n, prevSum, multi, add);}// Function to calculate and // return the Sum upto Nth termstatic int seriesSum(int n){ if (n == 1) return 0; int prevSum = 0; int multi = 1; int add = 1; // Recursive Function return seriesSumUtil(2, n, prevSum, multi, add);}// Driver codepublic static void Main(String []args){ int N = 5; Console.WriteLine(seriesSum(N)); }}// This code is contributed by Rohit_ranjan |
Javascript
<script>// Javascript program to calculate the// sum upto the Nth term of the// given series// Recursive Function to calculate the// sum upto Nth termfunction seriesSumUtil(k, n, prevSum, multi, add){ // If N-th term is calculated if (k == n + 1) { return prevSum; } // Update multi to store // product upto K multi = multi * k; // Update add to store // sum upto K add = add + k; // Update prevSum to store sum // upto K prevSum = prevSum + multi - add; // Proceed to next K return seriesSumUtil(k + 1, n, prevSum, multi, add);}// Function to calculate and// return the Sum upto Nth termfunction seriesSum(n){ if (n == 1) return 0; let prevSum = 0; let multi = 1; let add = 1; // Recursive Function return seriesSumUtil(2, n, prevSum, multi, add);}// Driver codelet N = 5;document.write(seriesSum(N));// This code is contributed by rameshtravel07</script> |
Output:
118
Time Complexity: O(N)
Auxiliary Space: O(N), for recursive stack space.
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