Given a positive number N, the task is to find the sum of the first (N + 1) Fibonacci Numbers.
Examples:
Input: N = 3
Output: 4
Explanation:
The first 4 terms of the Fibonacci Series is {0, 1, 1, 2}. Therefore, the required sum = 0 + 1 + 1 + 2 = 4.Input: N = 4
Output: 7
Naive Approach: For the simplest approach to solve the problem, refer to the previous post of this article.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by the following observations and calculations:
Let S(N) represent the sum of the first N terms of Fibonacci Series. Now, in order to simply find S(N), calculate the (N + 2)th Fibonacci number and subtract 1 from the result. The Nth term of this series can be calculated by the formula:
Now the value of S(N) can be calculated by (FN + 2 – 1).
Therefore, the idea is to calculate the value of SN using the above formula:
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of// first N + 1 fibonacci numbersvoid sumFib(int N){ // Apply the formula long num = (long)round(pow((sqrt(5) + 1) / 2.0, N + 2) / sqrt(5)); // Print the result cout << (num - 1);}// Driver Codeint main(){ int N = 3; sumFib(N); return 0;}// This code is contributed by Dharanendra L V. |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to find the sum of // first N + 1 fibonacci numbers public static void sumFib(int N) { // Apply the formula long num = (long)Math.round( Math.pow((Math.sqrt(5) + 1) / 2.0, N + 2) / Math.sqrt(5)); // Print the result System.out.println(num - 1); } // Driver Code public static void main(String[] args) { int N = 3; sumFib(N); }} |
Python3
# Python program for the above approachimport math# Function to find the sum of# first N + 1 fibonacci numbersdef sumFib(N): # Apply the formula num = round(pow((pow(5,1/2) + 1) \ / 2.0, N + 2) \ / pow(5,1/2)); # Print the result print(num - 1);# Driver Codeif __name__ == '__main__': N = 3; sumFib(N);# This code is contributed by 29AjayKumar |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the sum of // first N + 1 fibonacci numbers public static void sumFib(int N) { // Apply the formula long num = (long)Math.Round( Math.Pow((Math.Sqrt(5) + 1) / 2.0, N + 2) / Math.Sqrt(5)); // Print the result Console.WriteLine(num - 1); }// Driver Codestatic public void Main(){ int N = 3; sumFib(N);}}// This code is contributed by jana_sayantan. |
Javascript
<script>// Javascript program for the above approach // Function to find the sum of // first N + 1 fibonacci numbers function sumFib(N) { // Apply the formula var num = Math.round(Math.pow((Math.sqrt(5) + 1) / 2.0, N + 2) / Math.sqrt(5)); // Print the result document.write(num - 1); } // Driver Code var N = 3; sumFib(N);// This code contributed by umadevi9616</script> |
Output:
4
Time Complexity: O(log n)
Auxiliary Space: O(1)
Alternate Approach: Follow the steps below to solve the problem:
- The Nth Fibonacci number can also be calculated using the generating function.
- The generating function for the Nth Fibonacci number is given by:
- Therefore, the generating function of the sum of Fibonacci numbers is given by:
- After partial fraction decomposition of G'(X), the value of G'(X) is given by:
where,
- Therefore, the formula for the sum of the Fibonacci numbers is given by:
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of// first N + 1 fibonacci numbersvoid sumFib(int N){ // Apply the formula double num = (1 - sqrt(5)) / 2; long val = round( abs(1 / (pow(num, N + 2) + pow(num, N + 1) + pow(num, N) + pow(num, N - 1))) - 1); // Print the result cout<<val;}// Driver Codeint main() { int N = 3; // Function Call sumFib(N);}// This code is contributed by 29AjayKumar |
Java
// Java program for the above approachimport java.io.*;class GFG { // Function to find the sum of // first N + 1 fibonacci numbers public static void sumFib(int N) { // Apply the formula double num = (1 - Math.sqrt(5)) / 2; long val = Math.round( Math.abs(1 / (Math.pow(num, N + 2) + Math.pow(num, N + 1) + Math.pow(num, N) + Math.pow(num, N - 1))) - 1); // Print the result System.out.println(val); } // Driver Code public static void main(String[] args) { int N = 3; // Function Call sumFib(N); }} |
Python3
# Python3 program for the above approachimport math# Function to find the sum of# first N + 1 fibonacci numbersdef sumFib(N): # Apply the formula num = (1 - math.sqrt(5)) / 2 val = round(abs(1 / (pow(num, N + 2) + pow(num, N + 1) + pow(num, N) + pow(num, N - 1))) - 1) # Print the result print(val)# Driver Codeif __name__ == '__main__': N = 3 # Function Call sumFib(N)# This code is contributed by Surbhi Tyagi. |
C#
// C# program for the above approachusing System;public class GFG { // Function to find the sum of // first N + 1 fibonacci numbers public static void sumFib(int N) { // Apply the formula double num = (1 - Math.Sqrt(5)) / 2; double val = Math.Round( Math.Abs(1 / (Math.Pow(num, N + 2) + Math.Pow(num, N + 1) + Math.Pow(num, N) + Math.Pow(num, N - 1))) - 1); // Print the result Console.WriteLine(val); } // Driver Code public static void Main(String[] args) { int N = 3; // Function Call sumFib(N); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach // Function to find the sum of // first N + 1 fibonacci numbers function sumFib(N) { // Apply the formula var num = ((1 - Math.sqrt(5)) / 2); var val = Math.round( Math.abs(1 / (Math.pow(num, N + 2) + Math.pow(num, N + 1) + Math.pow(num, N) + Math.pow(num, N - 1))) - 1); // Print the result document.write(val); } // Driver Code var N = 3; // Function Call sumFib(N);// This code is contributed by todaysgaurav </script> |
Output:
4
Time Complexity: O(log n)
Auxiliary Space: O(1)
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