Given a natural number n, find the sum of the sum-series of the first N natural number.
Sum-Series: is sum of first N natural numbers, i.e, sum-series of 5 is 15 ( 1 + 2 + 3 + 4 + 5 ).
Natural number 1 2 3 4 5 6 Sum of natural number (sum-series) 1 3 6 10 15 21 Sum of sum-series 1 4 10 20 35 56
Example:
Input: N = 5
Output: 35
Explanation:
Sum of sum-series of {1, 2, 3, 4, 5} i.e. {1 + 3 + 6 + 10 + 15} is 35.Input: N = 2
Output: 4
Explanation:
Sum of sum-series of {1, 2} i.e. {1 + 3} is 4.
Simple approach:
Find sum series for every value from 1 to N and then add it.
- Create a variable Total_sum to store the required sum series.
- Iterate over the number from 1 to N
- Find sum-series of every value by using the formulae sum = (N*(N + 1)) / 2
- Add the value to Total_sum
- In the end, print the value stored in Total_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 find the sumstatic long sumOfSumSeries(int N){ long sum = 0L; // Calculate sum-series // for every natural number // and add them for (int i = 1; i <= N; i++) { sum = sum + (i * (i + 1)) / 2; } return sum;}// Driver codeint main(){ int N = 5; cout << sumOfSumSeries(N);}// This code is contributed by Code_Mech |
Java
// Java program to implement// the above approachclass GFG { // Function to find the sum static long sumOfSumSeries(int N) { long sum = 0L; // Calculate sum-series // for every natural number // and add them for (int i = 1; i <= N; i++) { sum = sum + (i * (i + 1)) / 2; } return sum; } // Driver code public static void main(String[] args) { int N = 5; System.out.println(sumOfSumSeries(N)); }} |
Python3
# Python3 program to implement # the above approach # Function to find the sum def sumOfSumSeries(N): _sum = 0 # Calculate sum-series # for every natural number # and add them for i in range(N + 1): _sum = _sum + (i * (i + 1)) // 2 return _sum # Driver code N = 5print(sumOfSumSeries(N)) # This code is contributed by divyamohan123 |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to find the sumstatic long sumOfSumSeries(int N){ long sum = 0L; // Calculate sum-series // for every natural number // and add them for(int i = 1; i <= N; i++) { sum = sum + (i * (i + 1)) / 2; } return sum;}// Driver codepublic static void Main(){ int N = 5; Console.Write(sumOfSumSeries(N));}}// This code is contributed by Nidhi_Biet |
Javascript
<script> // Javascript program to implement // the above approach // Function to find the sum function sumOfSumSeries(N) { let sum = 0; // Calculate sum-series // for every natural number // and add them for (let i = 1; i <= N; i++) { sum = sum + (i * (i + 1)) / 2; } return sum; } let N = 5; document.write(sumOfSumSeries(N)); // This code is contributed by suresh07.</script> |
35
Time complexity: O(N)
Auxiliary Space: O(1)
Efficient approach:
Total_sum of the above series can be calculated directly by using the below formulae:
where N is the natural number
Proof of the above formula:
Lets assume N = 5
- Then the sum is sum of all the below elements in the table, let’s call this “result”
| 1 | ||||
| 1 | 2 | |||
| 1 | 2 | 3 | ||
| 1 | 2 | 3 | 4 | |
| 1 | 2 | 3 | 4 | 5 |
let’s populate the empty cells with the same value in other columns, lets’s call this “totalSum“
| 1 | 2 | 3 | 4 | 5 |
| 1 | 2 | 3 | 4 | 5 |
| 1 | 2 | 3 | 4 | 5 |
| 1 | 2 | 3 | 4 | 5 |
| 1 | 2 | 3 | 4 | 5 |
As sum of N numbers is repeated N times
totalSum = N * [(N*(N + 1))/2]
populated data = (1 times * 2) + (2 times * 3) + (3 times * 4) + (4 times * 5)
= 1*2 + 2*3 + 3*4 ……… +(N-1)*N
=[(N-1) * (N) * (N+1)]/3
- Since,
result = totalSum – populatedData
= N * [(N*(N+1))/2] – [(N-1) * (N) * (N+1)]/3
= (N*(N+1)*(N+2))/6
- Therefore
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <iostream>#include <math.h>using namespace std;// Function to find the sumlong sumOfSumSeries(int n){ return (n * (n + 1) * (n + 2)) / 6;}// Driver codeint main (){ int N = 5; cout << sumOfSumSeries(N); return 0;}// This code is contributed// by shivanisinghss2110 |
Java
// Java program to implement// the above approachclass GFG { // Function to find the sum static long sumOfSumSeries(int n) { return (n * (n + 1) * (n + 2)) / 6; } // Driver code public static void main(String[] args) { int N = 5; System.out.println(sumOfSumSeries(N)); }} |
Python3
# Python3 program to implement # the above approach# Function to find the sum def sumOfSumSeries(n): return (n * (n + 1) * (n + 2)) // 6# Driver code N = 5print(sumOfSumSeries(N))# This code is contributed by divyamohan123 |
C#
// C# program to implement the // above approachusing System;class GFG{// Function to find the sumstatic long sumOfSumSeries(int n){ return (n * (n + 1) * (n + 2)) / 6;}// Driver codepublic static void Main(String[] args){ int N = 5; Console.Write(sumOfSumSeries(N));}}// This code is contributed by Ritik Bansal |
Javascript
<script> // Javascript program to implement // the above approach // Function to find the sum function sumOfSumSeries(n) { return (n * (n + 1) * (n + 2)) / 6; } let N = 5; document.write(sumOfSumSeries(N));</script> |
35
Time complexity: O(1), considering multiplication, addition & division takes constant time.
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
