# Sum of Fibonacci Numbers in a range

• Difficulty Level : Medium
• Last Updated : 25 Aug, 2021

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 the help of the Golden Ratio.

Fibonacci’s 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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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.`

## Python3

 `# 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`

## C#

 `// C# implementation 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 */`

## PHP

 ``

## Javascript

 ``
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 the 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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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`

## 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#

 `// 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`

## PHP

 ``

## Javascript

 ``
Output:
`50`

Time Complexity: O(1)
Auxiliary Space: O(1)

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