27 lines
565 B
Markdown
27 lines
565 B
Markdown
#Math #Probability
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# Statement
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For $n \geq r$, $n, r \in \mathbb{N}$:
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$$
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\sum _{i = r}^n {i \choose r} = {n + 1 \choose r + 1}
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$$
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# Proof
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Let us have a base case $n = r$:
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$$
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{r \choose r} = {r + 1 \choose r + 1} = 1
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$$
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Now suppose $\sum _{i = r}^n {i \choose r} = {n + 1 \choose r + 1}$ for a certain $n$:
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$$
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\sum _{i = r}^n {i \choose r} + {n + 1 \choose r} \\
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= {n + 1 \choose r + 1} + {n + 1 \choose r} \\
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= {n + 2 \choose r + 1}
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$$
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Since $n = r$ is true, and if a case is true for $n$, it is true for $n + 1$, this statement is true for all $n \geq r$. |