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

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1maybe 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"

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1the 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

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1maybe, 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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1there is a think a way to do it by thinking, however it is almost like saying \[\dbinom{n}{k}=\dbinom{n1}{k}+\dbinom{n1}{k1}\]

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

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2You 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.
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