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Let $F(n)$ be the nth Fibonacci number, where $F(1)=F(2)=1$ .

Which of the following statements is true?
Option: 1
$F(n)> n$ for all positive integers $n$ .

Option: 2
$F(n) < 2n$ for all positive integers $n$

Option: 3
$F(n) = 2n$ for some positive integers $n$ .

Option: 4
$F(n) > 2n$ for some positive integers $n$ .

Answers (1)

best_answer

We will use the principle of mathematical induction to solve this problem.

Given that $F(n)$ is the nth Fibonacci number, where $F(1)=F(2)=1$ .

The base case is when $n=1$ , then $F(1)=1$ , which is less than $2(1)$ .

$\therefore F(1) < 2 (1)$ is true.

Now assume that the statement is true for some natural number k.

$\therefore F(k) < 2 (k)$

Consider the $(k+1)^{th}$ Fibonacci number.

We need to show that $F(k+\ 1) < 2 (k+\ 1)$

By the definition of the Fibonacci sequence, we have:

$\therefore$ $F\left( k+1 \right)=F\left( k \right)+F\left( k-\ 1 \right)$

$\therefore F\left( k+1 \right)< 2\left( k \right)+F\left( k-\ 1 \right)$

We also know that by the induction hypothesis,

$F\left( k-\ 1 \right)< 2\left( k-\ 1 \right)$

Therefore, we have:

$F\left( k+\ 1 \right)< 2\left( k \right)+2\left( k-\ 1 \right)$

$\Rightarrow F\left( k+\ 1 \right)< \left(4 k-\ 2 \right)$

This inequality is true for $K> 1$ , and it implies that,

$F\left( k+\ 1 \right)< 2\left( k+\ 1 \right)$

Therefore, the statement is true for all positive integers n.

Hence, $F(n)< 2n$ for all positive integers n.

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