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Prove:
\[\binom{n1}{k1}
+\binom{n2}{k1}
+\binom{n3}{k1}
+....+\binom{k1}{k1}
=\binom{n}{k}\]
 one year ago
 one year ago
Prove: \[\binom{n1}{k1} +\binom{n2}{k1} +\binom{n3}{k1} +....+\binom{k1}{k1} =\binom{n}{k}\]
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.1
i will guess induction
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
induction? is there by chance another method?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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"
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
the last part will be some annoying algebra with factorials, but since you have a sum with \(nk\) 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
for base case it is probably easiest to start at \(n=2\)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
can it be a gemoetric progression?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
there is a think a way to do it by thinking, however it is almost like saying \[\dbinom{n}{k}=\dbinom{n1}{k}+\dbinom{n1}{k1}\]
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
hm..so induction is the only method? but i havent learnt induction to prove these kind of things..
 one year ago

joemath314159Best ResponseYou've already chosen the best response.2
You can just keep using:\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\begin{matrix}n1 \\ k\end{matrix}\right)+\left(\begin{matrix}n1 \\ k1\end{matrix}\right)\]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:\[\left(\begin{matrix}n1 \\ k\end{matrix}\right)=\left(\begin{matrix}n2 \\ k\end{matrix}\right)+\left(\begin{matrix}n2 \\ k1\end{matrix}\right)\]So altogether we have:\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\begin{matrix}n2 \\ k\end{matrix}\right)+\left(\begin{matrix}n2 \\ k1\end{matrix}\right)+\left(\begin{matrix}n1 \\ k1\end{matrix}\right)\]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 n2 on the top becomes k1.
 one year ago
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