Answer
See solution
Work Step by Step
We have the Fibonacci sequence \(\{F_k\}\) satisfying
\[
F_k \;=\; F_{k-1} \;+\; F_{k-2}
\quad\text{for all integers }k \ge 2,
\]
with typical initial conditions \(F_0=0,F_1=1\) (or sometimes \(F_1=1,F_2=1\)). Below are the requested explanations and equations.
---
## (a) Why \(F_{k+1} = F_k + F_{k-1}\)
This is just the same Fibonacci recurrence, but we shift the index from \(k\) to \(k+1\). Specifically, if the original recurrence is
\[
F_k \;=\; F_{k-1} + F_{k-2},
\]
then replacing \(k\) by \(k+1\) gives
\[
F_{(k+1)} \;=\; F_{(k+1)-1} + F_{(k+1)-2}
\;=\;
F_k + F_{k-1}.
\]
Hence \(F_{k+1} = F_k + F_{k-1}\) for \(k \ge 1\).
---
## (b) Express \(F_{k+2}\) in Terms of \(F_{k+1}\) and \(F_k\)
Again we use the same shifting idea but replace \(k\) by \(k+2\) in the original recurrence:
\[
F_{k+2}
\;=\; F_{(k+2)-1} + F_{(k+2)-2}
\;=\; F_{k+1} + F_k.
\]
---
## (c) Express \(F_{k+3}\) in Terms of \(F_{k+2}\) and \(F_{k+1}\)
Likewise, shifting the index by 3 in the original recurrence:
\[
F_{k+3}
\;=\; F_{(k+3)-1} + F_{(k+3)-2}
\;=\; F_{k+2} + F_{k+1}.
\]
These identities all follow directly from the original Fibonacci definition by substituting appropriate shifted indices.
