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  1. Base case is obvious
  2. Suppose true for $k$. Have tried direct substitution, i.e., replace $k$ with $k+1$ and then simplify; have also tried to use ${{n}\choose{k}} = {{n-1}\choose{k}} + {{n-1}\choose{k-1}}, n,k >1$.
N. F. Taussig
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ralph
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  • Assuming that you intend that $r \in {0,1,2,\cdots, n}$, what the answer of Unexpected Confusion is indicating is that instead of using induction across one row, consider the rule involved in constructing Pascal's Triangle, and use induction to go from one row to the next. For example, the fourth row of Pascal's triangle, can be expressed as $1 - 4 - 6 - 4 -1$ or $\binom{4}{0} - \binom{4}{1} - \binom{4}{2} - \binom{4}{3} - \binom{4}{4}.$ ...see next comment – user2661923 Nov 16 '21 at 06:09
  • If you assume that the $4$-th row is all integers, and you note that each element in the 5th row is the sum of the two (presumably integer) numbers above it... The minor gap in the analysis is in proving the rule of Pascal's Triangle construction, which is based on the easily provable algebraic conclusion that $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$ – user2661923 Nov 16 '21 at 06:12

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${^n\mathbb{C}_k}$ is $\frac{n!}{k!(n - k)!}$, right? And you have stated that ${^n\mathbb{C}_k}=$ ${^{n-1}\mathbb{C}_k}+$ ${^{n-1}\mathbb{C}_{k-1}}$. What does that tell you about its induction if you continue with the fraction form and the expansion above, condensing it in a sum?, for $n, k \in \mathbb{N},$ and $0 < k < n$

Also if you know, the expansion ${^n\mathbb{C}_k}=$ ${^{n-1}\mathbb{C}_k}+$ ${^{n-1}\mathbb{C}_{k-1}}$, is obtained from the pascals identity. Consider this when using induction.

Dstarred
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