For the sum $$\sum_i^n {n-i \choose i}$$
I evaluate it for $n=1,2,3,4,5$
For $n=1$ we have $$\sum_{i=0}^1 {1-i \choose i} = {1 \choose 0} + {0 \choose 1} = 1 + 0 = 1$$ For $n=2$ we have $$\sum_{i=0}^2 {2-i \choose i} = {2 \choose 0} + {1 \choose 1} + {0 \choose 2} = 1 + 1 + 0 = 2$$ For $n=3$ we have $$\sum_{i=0}^3 {3-i \choose i} = {3 \choose 0} + {2 \choose 1} + {1 \choose 2} + {0 \choose 3} = 1 + 2 + 0 + 0 = 3$$ For $n=4$ we have $$\sum_{i=0}^4 {4-i \choose i} = {4 \choose 0} + {3 \choose 1} + {2 \choose 2} + {1 \choose 3} + {0 \choose 4} = 1 + 3 + 1 + 0 + 0 = 5$$ For $n=5$ we have $$\sum_{i=0}^5 {5 - i \choose i} = {5 \choose 0} + {4 \choose 1} + {3 \choose 2} + {2 \choose 3} + {1 \choose 4} + {0 \choose 5} = 1 + 4 + 3 + 0 + 0 + 0 = 8$$
And we can see the pattern $$1,2,3,5,8,....$$ which makes me think about the Fibonacci sequence
One can also evaluate $n=0$ $$\sum_0^0 = {0 - i \choose i} = {0 \choose 0} = 1$$
And so we have $$1,1,2,3,5,8,....$$
Now to prove that it is really the Fibonacci sequence, Can I prove it by induction ?
By starting from $n =2$, we have $$\sum_{i=0}^n {n-i \choose i} = f(n) = f(n-1) + f(n-2)$$
Do we start with a base case $n=2$,
$$\sum_{i=0}^2 {2-i \choose i} = 2 = f(1) + f(0)$$
And so the base case works.
Now assume it works up to $n = k$ (Strong Mathematical Induction)
That is we assume it works up to $$\sum_{i=0}^k {k -i \choose i} = f(k-1) + f(k-2)$$
Now we consider $k + 1$, Now I am stuck here. Any suggestions ?
