Given a number N, the task is to find the sum of the below series till N terms.
1+(1+2)/2+(1+2+3)/3+… till N terms
Examples:
Input: N = 3
Output: 4.5Input: N = 4
Output: 7
Approach:
From the given series, find the formula for the Nth term:
1st term = 1 = 1
2nd term = (1+2)/2 = 1.5
3rd term = (1+2+3)/3 = 2
4th term = (1+2+3+4)/4 = 2.5
.
.
Nth term = N*(N +1)/(2*N) = (N+1)/2
Derivation:
The following series of steps can be used to derive the formula to find the sum of N terms-
The sequence-
1+(1+2)/2+(1+2+3)/3+… till N terms
can be written as-
1 +1.5 +2 +2.5 +3 +3.5 +4 +…till N terms
The above series is an Arithmetic Progression(AP) series. So, we can directly apply the formula of sum of N terms in AP.
The Sum of N terms of an AP can also be given by SN,
SN = N * [First term + Last term] / 2
SN = N * [2 * a + (N – 1) * d] / 2 -(1)
From the above equation, it is known that,
a(first term)=1, d(common difference) = 1.5 -1= 0.5
Substituting the values of a and d in the equation (1), we get-
SN = N * (2 + (N – 1) * 0.5) / 2
Illustration:
Input: N = 5
Output: 10
Explanation:
SN = 5 * [2 * 1 + (5 – 1) * 0.5] / 2
= 5 * (2 + 2) / 2
= 5 * 2
= 10
Below is the C++ program to implement the above approach:
C++
// C++ program to find the sum of the// series 1+(1+2)/2+(1+2+3)/3+...// till N terms#include <bits/stdc++.h>using namespace std;// Function to return the sum// upto N term of the seriesdouble sumOfSeries(int N){ return ((double)N * (2 + ((double)N - 1) * 0.5)) / 2;}// Driver Codeint main(){ // Get the value of N int N = 6; cout << sumOfSeries(N); return 0;} |
Java
// Java program to find the sum of the// series 1+(1+2)/2+(1+2+3)/3+...// till N termsimport java.util.*;public class GFG{ // Function to return the sum // upto N term of the series static double sumOfSeries(int N) { return ((double)N * (2 + ((double)N - 1) * 0.5)) / 2; } // Driver Code public static void main(String args[]) { // Get the value of N int N = 6; System.out.println(sumOfSeries(N)); }}// This code is contributed by Samim Hossain Mondal. |
Python
# Python program to find the sum of the# series 1+(1+2)/2+(1+2+3)/3+...# till N terms# Function to return the sum# upto N term of the seriesdef sumOfSeries(N): return (N * (2 + (N - 1) * 0.5) / 2)# Driver Code# Get the value of NN = 6print(sumOfSeries(N))# This code is contributed by Samim Hossain Mondal. |
C#
// C# program to find the sum of the// series 1+(1+2)/2+(1+2+3)/3+...// till N termsusing System;class GFG{ // Function to return the sum // upto N term of the series static double sumOfSeries(int N) { return ((double)N * (2 + ((double)N - 1) * 0.5)) / 2; } // Driver Code public static void Main() { // Get the value of N int N = 6; Console.Write(sumOfSeries(N)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach // Function to return the sum // upto N term of the series function sumOfSeries(N) { return (N * (2 + (N - 1) * 0.5)) / 2; } // Driver Code // Get the value of N let N = 6; document.write(sumOfSeries(N)); // This code is contributed by Potta Lokesh </script> |
13.5
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.
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