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Mimi_x3
Group Title
Prove:
\[\binom{n1}{k1}
+\binom{n2}{k1}
+\binom{n3}{k1}
+....+\binom{k1}{k1}
=\binom{n}{k}\]
 2 years ago
 2 years ago
Mimi_x3 Group Title
Prove: \[\binom{n1}{k1} +\binom{n2}{k1} +\binom{n3}{k1} +....+\binom{k1}{k1} =\binom{n}{k}\]
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i will guess induction
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
induction? is there by chance another method?
 2 years ago

satellite73 Group TitleBest 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"
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
for base case it is probably easiest to start at \(n=2\)
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
can it be a gemoetric progression?
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

satellite73 Group TitleBest 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 \[\dbinom{n}{k}=\dbinom{n1}{k}+\dbinom{n1}{k1}\]
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
hm..so induction is the only method? but i havent learnt induction to prove these kind of things..
 2 years ago

joemath314159 Group TitleBest 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.
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
well, thank you!
 2 years ago
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