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Mimi_x3 Group Title

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

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  1. satellite73 Group Title
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    i will guess induction

    • one year ago
  2. Mimi_x3 Group Title
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    induction? is there by chance another method?

    • one year ago
  3. satellite73 Group Title
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    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
  4. satellite73 Group Title
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    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

    • one year ago
  5. satellite73 Group Title
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    for base case it is probably easiest to start at \(n=2\)

    • one year ago
  6. Mimi_x3 Group Title
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    can it be a gemoetric progression?

    • one year ago
  7. Mimi_x3 Group Title
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    geometric*

    • one year ago
  8. satellite73 Group Title
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    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
  9. satellite73 Group Title
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    there is a think a way to do it by thinking, however it is almost like saying \[\dbinom{n}{k}=\dbinom{n-1}{k}+\dbinom{n-1}{k-1}\]

    • one year ago
  10. Mimi_x3 Group Title
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    hm..so induction is the only method? but i havent learnt induction to prove these kind of things..

    • one year ago
  11. joemath314159 Group Title
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    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.

    • one year ago
  12. Mimi_x3 Group Title
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    well, thank you!

    • one year ago
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