Given a number N, the task is to print the first N terms of the following series:
2 15 41 80 132 197 275 366 470 587…
Examples:
Input: N = 7
Output: 2 15 41 80 132 197 275
Input: N = 3
Output: 2 15 41
Approach: From the given series we can find the formula for Nth term:
1st term = 2
2nd term = 15 = 13 * 1 + 2
3rd term = 41 = 13 * 2 + 15 = 13 * 3 + 2
4th term = 80 = 13 * 3 + 41 = 13 * 6 + 2
5th term = 132 = 13 * 4 + 80 = 13 * 10 + 2
.
.
Nth term = (13 * N * (N – 1)) / 2 + 2
Therefore:
Nth term of the series
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 above approach:
C++
// C++ implementation to print the // given with the given Nth term#include "bits/stdc++.h"using namespace std;// Function to print the seriesvoid printSeries(int N){ int ith_term = 0; // Generate the ith term and for (int i = 1; i <= N; i++) { ith_term = (13 * i * (i - 1)) / 2 + 2; cout << ith_term << ", "; }}// Driver Codeint main(){ int N = 7; printSeries(N); return 0;} |
Java
// Java implementation to print the // given with the given Nth termimport java.util.*;class GFG{// Function to print the seriesstatic void printSeries(int N){ int ith_term = 0; // Generate the ith term and for(int i = 1; i <= N; i++) { ith_term = (13 * i * (i - 1)) / 2 + 2; System.out.print(ith_term + ", "); }}// Driver Codepublic static void main(String[] args){ int N = 7; printSeries(N);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation to print the# given with the given Nth term# Function to print the seriesdef printSeries(N): ith_term = 0; # Generate the ith term and for i in range(1, N + 1): ith_term = (13 * i * (i - 1)) / 2 + 2; print(int(ith_term), ", ", end = "");# Driver Codeif __name__ == '__main__': N = 7; printSeries(N); # This code is contributed by amal kumar choubey |
C#
// C# implementation to print the // given with the given Nth termusing System;class GFG{// Function to print the seriesstatic void printSeries(int N){ int ith_term = 0; // Generate the ith term and for(int i = 1; i <= N; i++) { ith_term = (13 * i * (i - 1)) / 2 + 2; Console.Write(ith_term + ", "); }}// Driver Codepublic static void Main(String[] args){ int N = 7; printSeries(N);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// javascript implementation to print the // given with the given Nth term// Function to print the seriesfunction printSeries( N){ let ith_term = 0; // Generate the ith term and for (let i = 1; i <= N; i++) { ith_term = (13 * i * (i - 1)) / 2 + 2; document.write( ith_term + ", "); }}// Driver Code let N = 7; printSeries(N); // This code is contributed by gauravrajput1</script> |
Output:
2, 15, 41, 80, 132, 197, 275,
Time complexity: O(n) because using a for loop
Auxiliary Space: O(1)
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