Question

A function f defined as 2 f(n) = f(n+2) + f(n+1) where n > 2 and f(1) = f(2) = -1. What is the value of f(1) + f(2) + f(3) + f(4) + …. f(2018)?​

Answer 1

Step-by-step explanation:

mplex n1, n2, temp;. printf("For 1st complex number \n");. printf("Enter real and imaginary part respectively:\n");. scanf("%f %f", &n1.real,

Answer 2

Given:

2 f(n) = f(n+2) + f(n+1) where n > 2

f(1) = f(2) = -1

To find:

Value of f(1) + f(2) + f(3) + f(4) + …. f(2018)

Solution:

From the equation given in the question,

When we consider n=1 ,2f(1) = f(3) + f(2)

-2 = f(3) - 1

f(3) = -1

When we take n=2

2f(2) = f(3) + f(4)

-2 = -1 + f(4)

f(4) = -1

If we determine the value like this till 2018th term, all the terms will result in a value -1.

As all the terms will be -1.

The sum of all the terms = -2018

Therefore, the sum of the given terms is -2018.