Sum of Fibonacci Numbers in a range

Given a range [l, r], the task is to find the sum fib(l) + fib(l + 1) + fib(l + 2) + ….. + fib(r) where fib(n) is the nth Fibonacci number.

Examples:

Input: l = 2, r = 5
Output: 11
fib(2) + fib(3) + fib(4) + fib(5) = 1 + 2 + 3 + 5 = 11

Input: l = 4, r = 8
Output: 50

Naive approach: Simply calculate fib(l) + fib(l + 1) + fib(l + 2) + ….. + fib(r) in O(r – l) time complexity.
In order to find fib(n) in O(1) we will take help of Golden Ratio.
Fibonacci calculation using Binet’s Formula

fib(n) = phin – psin) / ?5
Where,
phi = (1 + sqrt(5)) / 2 which is roughly equal to 1.61803398875
psi = 1 – phi = (1 – sqrt(5)) / 2 which is roughly equal to 0.61803398875

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
  
// Function to return the nth Fibonacci number
int fib(int n)
{
    double phi = (1 + sqrt(5)) / 2;
    return round(pow(phi, n) / sqrt(5));
}
  
// Function to return the required sum
ll calculateSum(int l, int r)
{
  
    // To store the sum
    ll sum = 0;
  
    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);
  
    return sum;
}
  
// Driver code
int main()
{
    int l = 4, r = 8;
    cout << calculateSum(l, r);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.lang.Math;
class GFG
{
      
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.sqrt(5)) / 2;
    return (int)Math.round(Math.pow(phi, n) / Math.sqrt(5));
}
  
// Function to return the required sum
static int calculateSum(int l, int r)
{
  
    // To store the sum
    int sum = 0;
  
    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);
  
    return sum;
}
  
// Driver code
public static void main(String[] args)
{
    int l = 4, r = 8;
    System.out.println(calculateSum(l, r));
  
}
}
  
//This code is contributed by Code_Mech.

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Python3

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# Python3 implementation of the approach
  
# Function to return the nth
# Fibonacci number
def fib(n):
    phi = ((1 + (5 ** (1 / 2))) / 2);
    return round((phi ** n) / (5 ** (1 / 2)));
  
# Function to return the required sum
def calculateSum(l, r):
      
    # To store the sum
    sum = 0;
  
    # Calculate the sum
    for i in range(l, r + 1):
        sum += fib(i);
  
    return sum;
  
# Driver Code
if __name__ == '__main__':
    l, r = 4, 8;
    print(calculateSum(l, r));
  
# This code contributed by Rajput-Ji

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C#

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// C# implemenatation of above approach 
using System;
  
class GFG
{
      
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.Sqrt(5)) / 2;
    return (int)Math.Round(Math.Pow(phi, n) / Math.Sqrt(5));
}
  
// Function to return the required sum
static int calculateSum(int l, int r)
{
  
    // To store the sum
    int sum = 0;
  
    // Calculate the sum
    for (int i = l; i <= r; i++)
        sum += fib(i);
  
    return sum;
}
  
// Driver code
public static void Main()
{
    int l = 4, r = 8;
    Console.WriteLine(calculateSum(l, r));
}
}
  
/* This code contributed by PrinciRaj1992 */

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PHP

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<?php
// PHP implementation of the approach
  
// Function to return the nth 
// Fibonacci number
function fib($n)
{
    $phi = (1 + sqrt(5)) / 2;
    return (int)round(pow($phi, $n) / sqrt(5));
}
  
// Function to return the required sum
function calculateSum($l, $r)
{
  
    // To store the sum
    $sum = 0;
  
    // Calculate the sum
    for ($i = $l; $i <= $r; $i++)
        $sum += fib($i);
  
    return $sum;
}
  
// Driver code
$l = 4;
$r = 8;
echo calculateSum($l, $r);
  
// This code is contributed by mits
?>

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Output:

50

Efficient approach: The idea is to find the relationship between the sum of Fibonacci numbers and nth Fibonacci number and use Binet’s Formula to calculate its value.

Relationship Deduction

  1. F(i) refers to the ith Fibonacci number.
  2. S(i) refers to sum of Fibonacci numbers till F(i).

We can rewrite the relation F(n + 1) = F(n) + F(n – 1) as below:
F(n – 1) = F(n + 1) – F(n)
Similarly,
F(n – 2) = F(n) – F(n – 1)



F(0) = F(2) – F(1)

Adding all the equations, on left side, we have
F(0) + F(1) + … + F(n – 1) which is S(n – 1)

Therefore,
S(n – 1) = F(n + 1) – F(1)
S(n – 1) = F(n + 1) – 1
S(n) = F(n + 2) – 1

In order to find S(n), simply calculate the (n + 2)th Fibonacci number and subtract 1 from the result.
Therefore,
S(l, r) = S(r) – S(l – 1)
S(l, r) = F(r + 2) – 1 – (F(l + 1) – 1)
S(l, r) = F(r + 2) – F(l + 1)

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the nth Fibonacci number
int fib(int n)
{
    double phi = (1 + sqrt(5)) / 2;
    return round(pow(phi, n) / sqrt(5));
}
  
// Function to return the required sum
int calculateSum(int l, int r)
{
  
    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}
  
// Driver code
int main()
{
    int l = 4, r = 8;
    cout << calculateSum(l, r);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
  
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.sqrt(5)) / 2;
    return (int) Math.round(Math.pow(phi, n) / Math.sqrt(5));
}
  
// Function to return the required sum
static int calculateSum(int l, int r)
{
  
    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}
  
// Driver code
public static void main(String[] args) 
{
    int l = 4, r = 8;
    System.out.println(calculateSum(l, r));
  
}
}
  
// This code is contributed by 29AjayKumar

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Python3

# Python3 implementation of the approach
import math

# Function to return the nth
# Fibonacci number
def fib(n):

phi = (1 + math.sqrt(5)) / 2;
return int(round(pow(phi, n) /
math.sqrt(5)));

# Function to return the required sum
def calculateSum(l, r):

# Using our deduced result
sum = fib(r + 2) – fib(l + 1);
return sum;

# Driver code
l = 4;
r = 8;
print(calculateSum(l, r));

# This code is contributed by mits

C#

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// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to return the nth Fibonacci number
static int fib(int n)
{
    double phi = (1 + Math.Sqrt(5)) / 2;
    return (int) Math.Round(Math.Pow(phi, n) / 
                            Math.Sqrt(5));
}
  
// Function to return the required sum
static int calculateSum(int l, int r)
{
  
    // Using our deduced result
    int sum = fib(r + 2) - fib(l + 1);
    return sum;
}
  
// Driver code
public static void Main() 
{
    int l = 4, r = 8;
    Console.WriteLine(calculateSum(l, r));
}
}
  
// This code is contributed
// by Akanksha Rai

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PHP

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<?php
// PHP implementation of the approach
  
// Function to return the nth Fibonacci number
function fib($n)
{
    $phi = (1 + sqrt(5)) / 2;
    return (int) round(pow($phi, $n) / sqrt(5));
}
  
// Function to return the required sum
function calculateSum($l, $r)
{
  
    // Using our deduced result
    $sum = fib($r + 2) - fib($l + 1);
    return $sum;
}
  
// Driver code
$l = 4; $r = 8;
echo(calculateSum($l, $r));
  
// This code is contributed by Code_Mech
?>

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Output:

50


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