Given a positive integer, N. Find the sum of the first N term of the series-
1/1*3, 1/3*5, 1/5*7, ….
Examples:
Input: N = 3
Output: 0.428571
Input: N = 1
Output: 0.333333
Approach: The sequence is formed by using the following pattern. For any value N-
SN = N / (2 * N + 1)
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 return sum of// N term of the seriesdouble findSum(int N) { return (double)N / (2 * N + 1); }// Driver Codeint main(){ int N = 3; cout << findSum(N);} |
Java
// JAVA program to implement// the above approachimport java.util.*;class GFG { // Function to return sum of // N term of the series public static double findSum(int N) { return (double)N / (2 * N + 1); } // Driver Code public static void main(String[] args) { int N = 3; System.out.print(findSum(N)); }}// This code is contributed by Taranpreet |
Python3
# Python 3 program for the above approach# Function to return sum of# N term of the seriesdef findSum(N): return N / (2 * N + 1)# Driver Codeif __name__ == "__main__": # Value of N N = 3 print(findSum(N))# This code is contributed by Abhishek Thakur. |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to return sum of // N term of the series public static double findSum(int N) { return (double)N / (2 * N + 1); } // Driver Code public static void Main() { int N = 3; Console.Write(findSum(N)); }}// This code is contributed by gfgking |
Javascript
<script>// Javascript program to implement// the above approach// Function to return sum of// N term of the seriesfunction findSum(N) { return N / (2 * N + 1); }// Driver Codelet N = 3;document.write(findSum(N));// This code is contributed by Palak Gupta</script> |
Output
0.428571
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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