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Did i prove the Binomial Theorem correctly? I got a feeling I did, but need another set of eyes to look over my work. Not really much of a question, sorry.

Binomial Theorem

$$(x+y)^{n}=\sum_{k=0}^{n}{{n}\choose{k}}x^{n-k}y^{k}$$

Base Case: $n=0$

$$(x+y)^{0}=1={{0}\choose{0}}x^{0-0}y^{0}=\sum_{k=0}^{0}{{0}\choose{k}}x^{0-k}y^{k}$$

Induction Hypothesis

$$(x+y)^{n}=\sum_{k=0}^{n}{{n}\choose{k}}x^{n-k}y^{k}$$

Induction Step

$$\begin{align*} (x+y)^{n+1} &= (x+y)(x+y)^{n} \\ &= x\sum_{k=0}^{n}{{n}\choose{k}}x^{n-k}y^{k}+y\sum_{k=0}^{n}{{n}\choose{k}}x^{n-k}y^{k} \\ &= \sum_{k=0}^{n}{{n}\choose{k}}x^{n+1-k}y^{k}+\sum_{k=0}^{n}{{n}\choose{k}}x^{n-k}y^{k+1} \\ &= {n\choose{0}}x^{n+1}+\sum_{k=1}^{n}{{n}\choose{k}}x^{n+1-k}y^{k}+{n\choose{n}}y^{n+1}+\sum_{k=0}^{n-1}{{n}\choose{k}}x^{n-k}y^{k+1} \\ &= x^{n+1}+y^{n+1}+\sum_{k=1}^{n}{{n}\choose{k}}x^{n+1-k}y^{k}+\sum_{k=0}^{n-1}{{n}\choose{k}}x^{n-k}y^{k+1} \\ &= {{n+1}\choose{0}}x^{n+1}+{{n+1}\choose{n+1}}y^{n+1}+\sum_{k=1}^{n}{{n}\choose{k}}x^{n+1-k}y^{k}+\sum_{k=1}^{n}{{n}\choose{k-1}}x^{n+1-k}y^{k} \\ &= {{n+1}\choose{0}}x^{n+1}+{{n+1}\choose{n+1}}y^{n+1}+\sum_{k=1}^{n}\left({{n}\choose{k}}+{{n}\choose{k-1}}\right)x^{n+1-k}y^{k} \\ &= {{n+1}\choose{0}}x^{n+1}+{{n+1}\choose{n+1}}y^{n+1}+\sum_{k=1}^{n}{{n+1}\choose{k}}x^{n+1-k}y^{k} \\ &= \sum_{k=0}^{n+1}{{n+1}\choose{k}}x^{n+1-k}y^{k}\end{align*}$$

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  • 1
    $\begingroup$ Please write your work in mathjax here, rather than including only a picture. There are also several proofs of this here on MSE, on Wikipedia, and in many discrete math textbooks. $\endgroup$ – user296602 Mar 13 '16 at 6:16
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    $\begingroup$ Hard on the eyes to proofread handwritten text. But everything looks right, the key is reindexing so you can use the Pascal Identity, which you did without an explicit reference, $\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}$. $\endgroup$ – André Nicolas Mar 13 '16 at 6:22

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