Sum of Fibonacci Numbers with alternate negatives
Last Updated :
25 Aug, 2022
Given a positive integer n, the task is to find the value of F1 - F2 + F3 -..........+ (-1)n+1Fn where Fi denotes i-th Fibonacci number.
Fibonacci Numbers: The Fibonacci numbers are the numbers in the following integer sequence.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ....
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
Fn = Fn-1 + Fn-2
with seed values F0 = 0 and F1 = 1.
Examples
Input: n = 5
Output: 4
Explanation: 1 - 1 + 2 - 3 + 5 = 4
Input: n = 8
Output: -12
Explanation: 1 - 1 + 2 - 3 + 5 - 8 + 13 - 21 = -12
Method 1: (O(n) time Complexity) This method includes solving the problem directly by finding all Fibonacci numbers till n and adding up the alternating sum. But this will require O(n) time complexity.
Below is the implementation of the above approach:
C++
// C++ Program to find alternate sum
// of Fibonacci numbers
#include <bits/stdc++.h>
using namespace std;
// Computes value of first fibonacci numbers
// and stores their alternate sum
int calculateAlternateSum(int n)
{
if (n <= 0)
return 0;
int fibo[n + 1];
fibo[0] = 0, fibo[1] = 1;
// Initialize result
int sum = pow(fibo[0], 2) + pow(fibo[1], 2);
// Add remaining terms
for (int i = 2; i <= n; i++) {
fibo[i] = fibo[i - 1] + fibo[i - 2];
// For even terms
if (i % 2 == 0)
sum -= fibo[i];
// For odd terms
else
sum += fibo[i];
}
// Return the alternating sum
return sum;
}
// Driver program to test above function
int main()
{
// Get n
int n = 8;
// Find the alternating sum
cout << "Alternating Fibonacci Sum upto "
<< n << " terms: "
<< calculateAlternateSum(n) << endl;
return 0;
}
// Java Program to find alternate sum
// of Fibonacci numbers
public class GFG {
//Computes value of first fibonacci numbers
// and stores their alternate sum
static double calculateAlternateSum(int n)
{
if (n <= 0)
return 0;
int fibo[] = new int [n + 1];
fibo[0] = 0;
fibo[1] = 1;
// Initialize result
double sum = Math.pow(fibo[0], 2) + Math.pow(fibo[1], 2);
// Add remaining terms
for (int i = 2; i <= n; i++) {
fibo[i] = fibo[i - 1] + fibo[i - 2];
// For even terms
if (i % 2 == 0)
sum -= fibo[i];
// For odd terms
else
sum += fibo[i];
}
// Return the alternating sum
return sum;
}
// Driver code
public static void main(String args[])
{
// Get n
int n = 8;
// Find the alternating sum
System.out.println("Alternating Fibonacci Sum upto "
+ n + " terms: "
+ calculateAlternateSum(n));
}
// This Code is contributed by ANKITRAI1
}
# Python 3 Program to find alternate sum
# of Fibonacci numbers
# Computes value of first fibonacci numbers
# and stores their alternate sum
def calculateAlternateSum(n):
if (n <= 0):
return 0
fibo = [0]*(n + 1)
fibo[0] = 0
fibo[1] = 1
# Initialize result
sum = pow(fibo[0], 2) + pow(fibo[1], 2)
# Add remaining terms
for i in range(2, n+1) :
fibo[i] = fibo[i - 1] + fibo[i - 2]
# For even terms
if (i % 2 == 0):
sum -= fibo[i]
# For odd terms
else:
sum += fibo[i]
# Return the alternating sum
return sum
# Driver program to test above function
if __name__ == "__main__":
# Get n
n = 8
# Find the alternating sum
print( "Alternating Fibonacci Sum upto "
, n ," terms: "
, calculateAlternateSum(n))
# this code is contributed by
# ChitraNayal
// C# Program to find alternate sum
// of Fibonacci numbers
using System;
class GFG
{
// Computes value of first fibonacci numbers
// and stores their alternate sum
static double calculateAlternateSum(int n)
{
if (n <= 0)
return 0;
int []fibo = new int [n + 1];
fibo[0] = 0;
fibo[1] = 1;
// Initialize result
double sum = Math.Pow(fibo[0], 2) +
Math.Pow(fibo[1], 2);
// Add remaining terms
for (int i = 2; i <= n; i++)
{
fibo[i] = fibo[i - 1] + fibo[i - 2];
// For even terms
if (i % 2 == 0)
sum -= fibo[i];
// For odd terms
else
sum += fibo[i];
}
// Return the alternating sum
return sum;
}
// Driver code
public static void Main()
{
// Get n
int n = 8;
// Find the alternating sum
Console.WriteLine("Alternating Fibonacci Sum upto " +
n + " terms: " + calculateAlternateSum(n));
}
}
// This code is contributed by inder_verma
<?php
// PHP Program to find alternate sum
// of Fibonacci numbers
// Computes value of first fibonacci
// numbers and stores their alternate sum
function calculateAlternateSum($n)
{
if ($n <= 0)
return 0;
$fibo = array();
$fibo[0] = 0;
$fibo[1] = 1;
// Initialize result
$sum = pow($fibo[0], 2) +
pow($fibo[1], 2);
// Add remaining terms
for ($i = 2; $i <= $n; $i++)
{
$fibo[$i] = $fibo[$i - 1] +
$fibo[$i - 2];
// For even terms
if ($i % 2 == 0)
$sum -= $fibo[$i];
// For odd terms
else
$sum += $fibo[$i];
}
// Return the alternating sum
return $sum;
}
// Driver Code
// Get n
$n = 8;
// Find the alternating sum
echo ("Alternating Fibonacci Sum upto ");
echo $n ;
echo " terms: ";
echo (calculateAlternateSum($n)) ;
// This code isw contributed
// by Shivi_Aggarwal
?>
<script>
// Javascript Program to find alternate sum
// of Fibonacci numbers
// Computes value of first fibonacci numbers
// and stores their alternate sum
function calculateAlternateSum(n)
{
if (n <= 0)
return 0;
var fibo = Array(n + 1).fill(0);
fibo[0] = 0;
fibo[1] = 1;
// Initialize result
var sum = Math.pow(fibo[0], 2) +
Math.pow(fibo[1], 2);
// Add remaining terms
for (i = 2; i <= n; i++) {
fibo[i] = fibo[i - 1] + fibo[i - 2];
// For even terms
if (i % 2 == 0)
sum -= fibo[i];
// For odd terms
else
sum += fibo[i];
}
// Return the alternating sum
return sum;
}
// Driver code
// Get n
var n = 8;
// Find the alternating sum
document.write(
"Alternating Fibonacci Sum upto " + n +" terms: "
+ calculateAlternateSum(n)
);
// This code contributed by gauravrajput1
</script>
Output: Alternating Fibonacci Sum upto 8 terms: -12
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 2: (O(log n) Complexity) This method involves the following observation to reduce the time complexity:
- For n = 2,
F1 - F2 = 1 - 1
= 0
= 1 + (-1)3 * F1 - For n = 3,
F1 - F2 + F3
= 1 - 1 + 2
= 2
= 1 + (-1)4 * F2 - For n = 4,
F1 - F2 + F3 - F4
= 1 - 1 + 2 - 3
= -1
= 1 + (-1)5 * F3 - For n = m,
F1 - F2 + F3 -.......+ (-1)m+1 * Fm-1
= 1 + (-1)m+1Fm-1
Assuming this to be true. Now if (n = m+1) is also true, it means that the assumption is correct. Otherwise, it is wrong. - For n = m+1,
F1 - F2 + F3 -.......+ (-1)m+1 * Fm + (-1)m+2 * Fm+1
= 1 + (-1)m+1 * Fm-1 + (-1)m+2 * Fm+1
= 1 + (-1)m+1(Fm-1 - Fm+1)
= 1 + (-1)m+1(-Fm) = 1 + (-1)m+2(Fm)
which is true as per the assumption for n = m.
Hence the general term for the alternating Fibonacci Sum:
F1 - F2 + F3 -.......+ (-1)n+1 Fn = 1 + (-1)n+1Fn-1
So in order to find an alternate sum, only the n-th Fibonacci term is to be found, which can be done in O(log n) time( Refer to Method 5 or 6 of this article.)
Below is the implementation of method 6 of this:
C++
// C++ Program to find alternate Fibonacci Sum in
// O(Log n) time.
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1000;
// Create an array for memoization
int f[MAX] = { 0 };
// Returns n'th Fibonacci number
// using table f[]
int fib(int n)
{
// Base cases
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already computed
if (f[n])
return f[n];
int k = (n & 1) ? (n + 1) / 2 : n / 2;
// Applying above formula [Note value n&1 is 1
// if n is odd, else 0].
f[n] = (n & 1) ? (fib(k) * fib(k) + fib(k - 1) * fib(k - 1))
: (2 * fib(k - 1) + fib(k)) * fib(k);
return f[n];
}
// Computes value of Alternate Fibonacci Sum
int calculateAlternateSum(int n)
{
if (n % 2 == 0)
return (1 - fib(n - 1));
else
return (1 + fib(n - 1));
}
// Driver program to test above function
int main()
{
// Get n
int n = 8;
// Find the alternating sum
cout << "Alternating Fibonacci Sum upto "
<< n << " terms : "
<< calculateAlternateSum(n) << endl;
return 0;
}
// Java Program to find alternate
// Fibonacci Sum in O(Log n) time.
class GFG {
static final int MAX = 1000;
// Create an array for memoization
static int f[] = new int[MAX];
// Returns n'th Fibonacci number
// using table f[]
static int fib(int n) {
// Base cases
if (n == 0) {
return 0;
}
if (n == 1 || n == 2) {
return (f[n] = 1);
}
// If fib(n) is already computed
if (f[n] > 0) {
return f[n];
}
int k = (n % 2 == 1) ? (n + 1) / 2 : n / 2;
// Applying above formula [Note value n&1 is 1
// if n is odd, else 0].
f[n] = (n % 2 == 1) ? (fib(k) * fib(k) + fib(k - 1) * fib(k - 1))
: (2 * fib(k - 1) + fib(k)) * fib(k);
return f[n];
}
// Computes value of Alternate Fibonacci Sum
static int calculateAlternateSum(int n) {
if (n % 2 == 0) {
return (1 - fib(n - 1));
} else {
return (1 + fib(n - 1));
}
}
// Driver program to test above function
public static void main(String[] args) {
// Get n
int n = 8;
// Find the alternating sum
System.out.println("Alternating Fibonacci Sum upto "
+ n + " terms : "
+ calculateAlternateSum(n));
}
}
// This code is contributed by PrinciRaj1992
# Python3 program to find alternative
# Fibonacci Sum in O(Log n) time.
MAX = 1000
# List for memorization
f = [0] * MAX
# Returns n'th fibonacci number
# using table f[]
def fib(n):
# Base Cases
if(n == 0):
return(0)
if(n == 1 or n == 2):
f[n] = 1
return(f[n])
# If fib(n) is already computed
if(f[n]):
return(f[n])
if(n & 1):
k = (n + 1) // 2
else:
k = n // 2
# Applying above formula [Note value n&1 is 1
# if n is odd, else 0]
if(n & 1):
f[n] = (fib(k) * fib(k) +
fib(k - 1) * fib(k - 1))
else:
f[n] = (2 * fib(k-1) + fib(k)) * fib(k)
return(f[n])
# Computes value of Alternate Fibonacci Sum
def cal(n):
if(n % 2 == 0):
return(1 - fib(n - 1))
else:
return(1 + fib(n - 1))
# Driver Code
if(__name__=="__main__"):
n = 8
print("Alternating Fibonacci Sum upto",
n, "terms :", cal(n))
# This code is contributed by arjunsaini9081
// C# Program to find alternate
// Fibonacci Sum in O(Log n) time.
using System;
class GFG
{
static readonly int MAX = 1000;
// Create an array for memoization
static int []f = new int[MAX];
// Returns n'th Fibonacci number
// using table f[]
static int fib(int n)
{
// Base cases
if (n == 0)
{
return 0;
}
if (n == 1 || n == 2)
{
return (f[n] = 1);
}
// If fib(n) is already computed
if (f[n] > 0)
{
return f[n];
}
int k = (n % 2 == 1) ? (n + 1) / 2 : n / 2;
// Applying above formula [Note value n&1 is 1
// if n is odd, else 0].
f[n] = (n % 2 == 1) ? (fib(k) * fib(k) +
fib(k - 1) * fib(k - 1))
: (2 * fib(k - 1) + fib(k)) * fib(k);
return f[n];
}
// Computes value of Alternate Fibonacci Sum
static int calculateAlternateSum(int n)
{
if (n % 2 == 0)
{
return (1 - fib(n - 1));
} else
{
return (1 + fib(n - 1));
}
}
// Driver code
public static void Main()
{
// Get n
int n = 8;
// Find the alternating sum
Console.WriteLine("Alternating Fibonacci Sum upto "
+ n + " terms : "
+ calculateAlternateSum(n));
}
}
// This code is contributed by 29AjayKumar
<script>
//Javascript implementation of the approach
MAX = 1000;
// Create an array for memoization
f = new Array(MAX);
f.fill(0);
// Returns n'th Fibonacci number
// using table f[]
function fib( n)
{
// Base cases
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already computed
if (f[n])
return f[n];
var k = (n & 1) ? (n + 1) / 2 : n / 2;
// Applying above formula [Note value n&1 is 1
// if n is odd, else 0].
f[n] = (n & 1) ? (fib(k) * fib(k) + fib(k - 1) * fib(k - 1))
: (2 * fib(k - 1) + fib(k)) * fib(k);
return f[n];
}
// Computes value of Alternate Fibonacci Sum
function calculateAlternateSum(n)
{
if (n % 2 == 0)
return (1 - fib(n - 1));
else
return (1 + fib(n - 1));
}
// Get n
var n = 8;
// Find the alternating sum
document.write( "Alternating Fibonacci Sum upto "+ n + " terms : "
+ calculateAlternateSum(n) + "<br>");
getElement(a, n, S);
// This code is contributed by SoumikMondal
</script>
Output: Alternating Fibonacci Sum upto 8 terms: -12
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