Fibonacci numbers

[mathworker]

we all know Fibonacci numbers,just for information
they are the numbers of sequence whose \(\displaystyle t_n=t_{n-1}+t_{n-2}\) and \(\displaystyle t_0=t_1=1\)

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

 

[MarkFL]

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:

\(\displaystyle 1+S_0=F_{0+2}\)

\(\displaystyle 1=1\)

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

\(\displaystyle 1+S_k=F_{k+2}\)

Add \(\displaystyle F_{k+1}\) to both sides:

\(\displaystyle 1+S_k+F_{k+1}=F_{k+2}+F_{k+1}\)

\(\displaystyle 1+S_{k+1}=F_{(k+1)+2}\)

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

 

[Chris L T521]

Re: fibonacci numbers

we all know Fibonacci numbers,just for information
they are the numbers of sequence whose \(\displaystyle t_n=t_{n-1}+t_{n-2}\) and \(\displaystyle t_0=t_1=1\)

\(\displaystyle \text{PROVE THAT:}\)
$$1+S_n=t_{n+2}$$
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, $S_n=t_{n+2}-1\implies \boxed{1+S_n=t_{n+2}}$