Given the first two terms of the series as 1 and 6 and all the elements of the series are 2 less than the mean of the number preceding and succeeding it. The task is to print the nth term of the series.
First few terms of the series are:
1, 6, 15, 28, 45, 66, 91, …
Examples:
Input: N = 3
Output: 15
Input: N = 1
Output: 1
Approach: The given series represents odd positioned numbers in the triangular number series. Since the nth triangular number can easily be found by (n * (n + 1) / 2), so for finding the odd numbers we can replace n by (2 * n) – 1 as (2 * n) – 1 will always result in odd numbers i.e. the nth number of the given series will be ((2 * n) – 1) * n.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the nth term// of the given seriesint oddTriangularNumber(int N){ return (N * ((2 * N) - 1));}// Driver codeint main(){ int N = 3; cout << oddTriangularNumber(N); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the nth term// of the given seriesstatic int oddTriangularNumber(int N){ return (N * ((2 * N) - 1));}// Driver codepublic static void main(String[] args) { int N = 3; System.out.println(oddTriangularNumber(N));}}// This code contributed by Rajput-Ji |
Python3
# Python 3 implementation of the approach# Function to return the nth term# of the given seriesdef oddTriangularNumber(N): return (N * ((2 * N) - 1))# Driver codeif __name__ == '__main__': N = 3 print(oddTriangularNumber(N))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approach using System;class GFG { // Function to return the nth term // of the given series static int oddTriangularNumber(int N) { return (N * ((2 * N) - 1)); } // Driver code public static void Main(String[] args) { int N = 3; Console.WriteLine(oddTriangularNumber(N)); } } /* This code contributed by PrinciRaj1992 */ |
PHP
<?php// PHP implementation of the approach // Function to return the nth term // of the given series function oddTriangularNumber($N) { return ($N * ((2 * $N) - 1)); } // Driver code $N = 3; echo oddTriangularNumber($N); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the nth term// of the given seriesfunction oddTriangularNumber(N){ return (N * ((2 * N) - 1));}// Driver codelet N = 3;document.write(oddTriangularNumber(N));// This code is contributed by subham348.</script> |
15
Time Complexity: O(1)
Auxiliary Space: O(1)
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