Program to print the series 1, 3, 4, 8, 15, 27, 50… till N terms
Given a number N, the task is to print the first N terms of the following series:
1, 3, 4, 8, 15, 27, 50…
Examples:
Input: N = 7
Output: 1, 3, 4, 8, 15, 27, 50
Input: N = 3
Output: 1, 3, 4
Approach: From the given series we can find the formula for Nth term:
1st term = 1, 2nd term = 3, 3rd term = 4
4th term = 1st term + 2nd term + 3rd term
5th term = 2nd term + 3rd term + 4th term
6th term = 3rd term + 4th term + 5th term
.
.
so on
Therefore, the idea is to keep track of the last three terms of the series and find the consecutive terms of the series.
Below is the implementation of above approach:
C++
// C++ implementation to print the // N terms of the series whose three // terms are given #include "bits/stdc++.h" using namespace std; // Function to print the series void printSeries( int n, int a, int b, int c) { int d; // Generate the ith term and // print it if (n == 1) { cout << a << " " ; return ; } if (n == 2) { cout << a << " " << b << " " ; return ; } cout << a << " " << b << " " << c << " " ; for ( int i = 4; i <= n; i++) { d = a + b + c; cout << d << " " ; a = b; b = c; c = d; } } // Driver Code int main() { int N = 7, a = 1, b = 3; int c = 4; // Function Call printSeries(N, a, b, c); return 0; } |
Java
// Java implementation to print the // N terms of the series whose three // terms are given //include "bits/stdJava.h" import java.util.*; class GFG{ // Function to print the series static void printSeries( int n, int a, int b, int c) { int d; // Generate the ith term and // print it if (n == 1 ) { System.out.print(a + " " ); return ; } if (n == 2 ) { System.out.print(a + " " + b + " " ); return ; } System.out.print(a + " " + b + " " + c + " " ); for ( int i = 4 ; i <= n; i++) { d = a + b + c; System.out.print(d + " " ); a = b; b = c; c = d; } } // Driver Code public static void main(String[] args) { int N = 7 , a = 1 , b = 3 ; int c = 4 ; // Function Call printSeries(N, a, b, c); } } // This code is contributed by sapnasingh4991 |
Python3
# Python3 implementation to print the # N terms of the series whose three # terms are given # Function to print the series def printSeries(n, a, b, c): # Generate the ith term and # print it if (n = = 1 ): print (a, end = " " ); return ; if (n = = 2 ): print (a, b, end = " " ); return ; print (a, b, c, end = " " ); for i in range ( 4 , n + 1 ): d = a + b + c; print (d, end = " " ); a = b; b = c; c = d; # Driver Code N = 7 ; a = 1 ; b = 3 ; c = 4 ; # Function Call printSeries(N, a, b, c); # This code is contributed by Code_Mech |
C#
// C# implementation to print the // N terms of the series whose three // terms are given using System; class GFG{ // Function to print the series static void printSeries( int n, int a, int b, int c) { int d; // Generate the ith term and // print it if (n == 1) { Console.Write(a + " " ); return ; } if (n == 2) { Console.Write(a + " " + b + " " ); return ; } Console.Write(a + " " + b + " " + c + " " ); for ( int i = 4; i <= n; i++) { d = a + b + c; Console.Write(d + " " ); a = b; b = c; c = d; } } // Driver Code public static void Main() { int N = 7, a = 1, b = 3; int c = 4; // Function call printSeries(N, a, b, c); } } // This code is contributed by rock cool |
Javascript
<script> // javascript implementation to print the // N terms of the series whose three // terms are given // Function to print the series function printSeries( n, a, b, c) { let d; // Generate the ith term and // print it if (n == 1) { document.write( a + " " ); return ; } if (n == 2) { document.write( a + " " + b + " " ); return ; } document.write( a + " " + b + " " + c + " " ); for (let i = 4; i <= n; i++) { d = a + b + c; document.write( d + " " ); a = b; b = c; c = d; } } // Driver Code let N = 7, a = 1, b = 3; let c = 4; // Function Call printSeries(N, a, b, c); // This code is contributed by gauravrajput1 </script> |
Output:
1 3 4 8 15 27 50
Time complexity: O(n) because using a for loop
Auxiliary Space: O(1), since no extra space has been taken.
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