Given a number N, the task is to print the first N terms of the series:
Examples:
Input: N = 7
Output: 1, 9, 17, 33, 49, 73, 97
Input: N = 3
Output: 1, 9, 17
Approach: From the given series, find the formula for Nth term:
1st term = 1 2nd term = 9 = 2 * 4 + 1 3rd term = 17 = 2 * 9 - 1 4th term = 33 = 2 * 16 + 1 5th term = 49 = 2 * 25 - 1 6th term = 73 = 2 * 36 + 1 . . Nth term = (2 * N2 + (-1)N)
Therefore:
Nth term of the series
*** QuickLaTeX cannot compile formula: *** Error message: Error: Nothing to show, formula is empty
Then iterate over numbers in the range [1, N] to find all the terms using the above formula and print them.
Below is the implementation of the above approach:
CPP
// C++ implementation of the above approach#include "bits/stdc++.h"using namespace std;// Function to print the seriesvoid printSeries(int N){ int ith_term = 0; // Generate the ith term and // print it for (int i = 1; i <= N; i++) { ith_term = i % 2 == 0 ? 2 * i * i + 1 : 2 * i * i - 1; cout << ith_term << ", "; }}// Driver Codeint main(){ int N = 7; printSeries(N); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;class GFG{ // Function to print the seriesstatic void printSeries(int N){ int ith_term = 0; // Generate the ith term and // print it for (int i = 1; i <= N; i++) { ith_term = i % 2 == 0 ? 2 * i * i + 1 : 2 * i * i - 1; System.out.print(ith_term+ ", "); }} // Driver Codepublic static void main(String[] args){ int N = 7; printSeries(N);}}// This code is contributed by PrinciRaj1992 |
Python3
# Python implementation of the above approach# Function to print seriesdef printSeries(N): ith_term = 0; # Generate the ith term and # print for i in range(1,N+1): ith_term = 0; if(i % 2 == 0): ith_term = 2 * i * i + 1; else: ith_term = 2 * i * i - 1; print(ith_term,end= ", "); # Driver Codeif __name__ == '__main__': N = 7; printSeries(N); # This code is contributed by Princi Singh |
C#
// C# implementation of the above approachusing System;class GFG{// Function to print the seriesstatic void printSeries(int N){ int ith_term = 0; // Generate the ith term and // print it for (int i = 1; i <= N; i++) { ith_term = i % 2 == 0? 2 * i * i + 1: 2 * i * i - 1; Console.Write(ith_term+ ", "); }}// Driver Codepublic static void Main(){ int N = 7; printSeries(N);}}// This code is contributed by AbhiThakur |
Javascript
<script>// javascript implementation of the above approach// Function to print the seriesfunction printSeries( N){ let ith_term = 0; // Generate the ith term and // print it for (let i = 1; i <= N; i++) { ith_term = i % 2 == 0 ? 2 * i * i + 1 : 2 * i * i - 1; document.write( ith_term + ", "); }}// Driver Code let N = 7; printSeries(N); // This code is contributed by gauravrajput1</script> |
Output:
1, 9, 17, 33, 49, 73, 97,
Time Complexity: O(N)
Auxiliary Space: O(1), since no extra space has been taken.
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