MHB Proving the Formula of Fibonacci Numbers

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The discussion focuses on proving the relationship between Fibonacci numbers and their sums, specifically that 1 plus the sum of the first n Fibonacci numbers equals the (n+2)th Fibonacci number. The proof is approached using mathematical induction, establishing a base case and an induction hypothesis. An alternative method without induction is also presented, utilizing the properties of Fibonacci numbers to derive the same conclusion. Both methods confirm that the formula holds true for Fibonacci sequences. The conclusion is that the relationship 1 + S_n = t_{n+2} is valid, where S_n is the sum of the first n terms.
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we all know Fibonacci numbers,just for information
they are the numbers of sequence whose $$t_n=t_{n-1}+t_{n-2}$$ and $$t_0=t_1=1$$

$$\text{PROVE THAT:}$$
$$1+S_n=t_{n+2}$$
where,
$$S_n=\text{sum up-to n terms}$$
$$t_n=\text{nth term}$$
 
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Re: fibonacci numbers

We can use induction here (I prefer the notation $F_n$ for the $n$th Fibonacci number). The base case $P_0$ is:

$$1+S_0=F_{0+2}$$

$$1=1$$

This is true. The induction hypothesis $P_k$ is then:

$$1+S_k=F_{k+2}$$

Add $$F_{k+1}$$ to both sides:

$$1+S_k+F_{k+1}=F_{k+2}+F_{k+1}$$

$$1+S_{k+1}=F_{(k+1)+2}$$

We have derived $P_{k+1}$ from $P_{k}$ thereby completing the proof by induction.
 
Re: fibonacci numbers

mathworker said:
we all know Fibonacci numbers,just for information
they are the numbers of sequence whose $$t_n=t_{n-1}+t_{n-2}$$ and $$t_0=t_1=1$$

$$\text{PROVE THAT:}$$
$$1+S_n=t_{n+2}$$
where,
$$S_n=\text{sum up-to n terms}$$
$$t_n=\text{nth term}$$

I'm going to do this without induction. The relation used to rewrite the terms in the sum will be $t_k=t_{k+2}-t_{k+1}$ for $k=0,\ldots,n$.

We have that
\[\begin{aligned}S_n &= t_n + t_{n-1}+t_{n-2}+\ldots+t_3+t_2+t_1+t_0\\ &= \underbrace{(t_{n+2}-t_{n+1})}_{t_n} + \underbrace{(t_{n+1}-t_n)}_{t_{n-1}} +\underbrace{(t_n-t_{n-1})}_{t_{n-2}} + \underbrace{(t_{n-1}-t_{n-2})}_{t_{n-3}} +\ldots+\underbrace{(t_5-t_4)}_{t_3} + \underbrace{(t_4-t_3)}_{t_2} + \underbrace{(t_3-t_2)}_{t_1}+\underbrace{(t_2-t_1)}_{t_0} \\ &= t_{n+2} -t_1\\ &= t_{n+2}-1\end{aligned}\]
Thus, $S_n=t_{n+2}-1\implies \boxed{1+S_n=t_{n+2}}$
 
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