Prove:
$$\binom{n-1}{k-1}
+\binom{n-2}{k-1}
+\binom{n-3}{k-1}
+....+\binom{k-1}{k-1}
=\binom{n}{k}$$

Question

Prove: 
$$\binom{n-1}{k-1}
 +\binom{n-2}{k-1}
 +\binom{n-3}{k-1}
 +....+\binom{k-1}{k-1}
 =\binom{n}{k}$$

anonymous · Answer

i  will guess induction

mimi_x3 · Answer

induction?
is there by chance another method?

anonymous · Answer

maybe a thinking method, but i can't imagine there is another way, since you are making a statement that says essentially "for all $n\in \mathbb{N}$ this formula holds"

anonymous · Answer

the last part will be some annoying algebra with factorials, but since you have a sum with $n-k$ terms you cannot add them i in a formula that i can think of.
i think the induction step will be enough arithmetic as it is

anonymous · Answer

for base case it is probably easiest to start at $n=2$

mimi_x3 · Answer

can it be a gemoetric progression?

mimi_x3 · Answer

geometric*

anonymous · Answer

maybe, lets try it with numbers , say $n=5, k=3$ then $\dbinom{2}{2}=1,\dbinom{3}{2}=3, \dbinom{4}{2}=6$ 
no, not geometric

anonymous · Answer

there is a think a way to do it by thinking, however
it is almost like saying $$\dbinom{n}{k}=\dbinom{n-1}{k}+\dbinom{n-1}{k-1}$$

mimi_x3 · Answer

hm..so induction is the only method?

but i havent learnt induction to prove these kind of things..

anonymous · Answer

You can just keep using:$$\left(\begin{matrix}n \ k\end{matrix}\right)=\left(\begin{matrix}n-1 \ k\end{matrix}\right)+\left(\begin{matrix}n-1 \ k-1\end{matrix}\right)$$over and over again.  In a sense, its an inductive argument. Using it once gives you the above statement. Notice the second part of the sum is something you want, but not the first part. So use that idea on the first one again:$$\left(\begin{matrix}n-1 \ k\end{matrix}\right)=\left(\begin{matrix}n-2 \ k\end{matrix}\right)+\left(\begin{matrix}n-2 \ k-1\end{matrix}\right)$$So altogether we have:$$\left(\begin{matrix}n \ k\end{matrix}\right)=\left(\begin{matrix}n-2 \ k\end{matrix}\right)+\left(\begin{matrix}n-2 \ k-1\end{matrix}\right)+\left(\begin{matrix}n-1 \ k-1\end{matrix}\right)$$Again, the last two parts of the sum are what you want, but the first isnt, so keep repeating this process (this is the inductive reasoning).  You do this all the way until the n-2 on the top becomes k-1.

mimi_x3 · Answer

well, thank you!