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Using induction\\ $\sum_{k=k}^{m} \binom{m}{k} = \frac{(n+1)!}{(k+1)!(n-k+1)!}$ \\ or any other method show that $\forall n, m \in \mathbb{N}$ \\ $\binom{m}{k} = \frac{m!}{k!(m-k)!}$ \\ $\binom{m}{k} = \frac{m!}{k!(m-k)!} \cdot \frac{(k+1)!(m+1-k)!}{(k+1)!(m+1-k)!} = \frac{(m+1)!}{(k+1)!(m+1-k)!}$

          Using induction\\
$\sum_{k=k}^{m} \binom{m}{k} = \frac{(n+1)!}{(k+1)!(n-k+1)!}$ \\
or any other method show that $\forall n, m \in \mathbb{N}$ \\
$\binom{m}{k} = \frac{m!}{k!(m-k)!}$ \\
$\binom{m}{k} = \frac{m!}{k!(m-k)!} \cdot \frac{(k+1)!(m+1-k)!}{(k+1)!(m+1-k)!} = \frac{(m+1)!}{(k+1)!(m+1-k)!}$

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Using induction

∑k=k^mmk = ((n+1)!)/((k+1)!(n-k+1)!)

or any other method show that ∀ n, m ∈ℕ

mk = (m!)/(k!(m-k)!)

mk = (m!)/(k!(m-k)!)·((k+1)!(m+1-k)!)/((k+1)!(m+1-k)!) = ((m+1)!)/((k+1)!(m+1-k)!)

Added by Alyssa R.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Anna D.

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