I've been trying to understand the proof for the binomial theorem

[Chenkel]

Hello everyone,

I've been trying to understand the proof for the binomial theorem and have been using this inductive proof for understanding.

So far the proof seems consistent everywhere it's explicit with the pattern it states, but I've started wondering if I actually fully grock it because I haven't seen all patterns of terms that might be possible.

I have trust that the binomial theorem works, I just want to be sure I fully understand the proof.

Let me know what you think, thank you!

 

[FactChecker]

but I've started wondering if I actually fully grock it because I haven't seen all patterns of terms that might be possible.

I'm not sure what you mean here. Do you mean that you want the '...' in the proof filled in? That would be impossible, although you might fill in a couple more yourself to convince yourself that the proof is legitimate. Other than that, I don't think there are any other patterns of terms to consider.

 

[PeroK]

Post to invoke Latex!

 

[PeroK]

Hello everyone,

I've been trying to understand the proof for the binomial theorem and have been using this inductive proof for understanding.

So far the proof seems consistent everywhere it's explicit with the pattern it states, but I've started wondering if I actually fully grock it because I haven't seen all patterns of terms that might be possible.

I have trust that the binomial theorem works, I just want to be sure I fully understand the proof.

Let me know what you think, thank you!

An alternative for the inductive proof is:
$$(a + b)^{n+1} = (a+b)(a+b)^n = (a + b)\sum_{k = 0}^n \binom n k a^{n-k}b^k$$$$= \sum_{k = 0}^n \binom n k a^{n-k+1}b^k+ \sum_{k = 0}^n \binom n k a^{n-k}b^{k+1}$$$$=\sum_{k = 0}^n \binom n k a^{n+1 -k}b^k+ \sum_{k = 1}^{n+1} \binom n {k-1} a^{n+1-k}b^{k}$$$$= a^{n+1} + \sum_{k = 1}^n \binom n k a^{n+1 -k}b^k+ \sum_{k = 1}^{n} \binom n {k-1} a^{n+1-k}b^{k} + b^{n+1}$$$$= a^{n+1} + \sum_{k = 1}^n \bigg [\binom n k +\binom n {k-1}\bigg] a^{n+1 -k}b^k + b^{n+1}$$$$= a^{n+1} + \sum_{k = 1}^n \binom {n+1} k a^{n+1 -k}b^k + b^{n+1}$$$$= \sum_{k = 0}^{n+1} \binom {n+1} k a^{n+1 -k}b^k$$

 

[PeroK]

PS the proof could be simplified somewhat by noting that it is sufficient to prove the identity when $a = 1$. Then, as a corollary we have:
$$(a + b)^n = a^n (1 + \frac b a)^n = a^n \sum_{k=0}^n \binom n k \big (\frac b a)^k$$$$= a^n \sum_{k=0}^n \binom n k a^{-k}b^k = \sum_{k=0}^n \binom n k a^{n-k}b^k$$

 

[Chenkel]

An alternative for the inductive proof is:
$$(a + b)^{n+1} = (a+b)(a+b)^n = (a + b)\sum_{k = 0}^n \binom n k a^{n-k}b^k$$$$= \sum_{k = 0}^n \binom n k a^{n-k+1}b^k+ \sum_{k = 0}^n \binom n k a^{n-k}b^{k+1}$$$$=\sum_{k = 0}^n \binom n k a^{n+1 -k}b^k+ \sum_{k = 1}^{n+1} \binom n {k-1} a^{n+1-k}b^{k}$$$$= a^{n+1} + \sum_{k = 1}^n \binom n k a^{n+1 -k}b^k+ \sum_{k = 1}^{n} \binom n {k-1} a^{n+1-k}b^{k} + b^{n+1}$$$$= a^{n+1} + \sum_{k = 1}^n \bigg [\binom n k +\binom n {k-1}\bigg] a^{n+1 -k}b^k + b^{n+1}$$$$= a^{n+1} + \sum_{k = 1}^n \binom {n+1} k a^{n+1 -k}b^k + b^{n+1}$$$$= \sum_{k = 0}^{n+1} \binom {n+1} k a^{n+1 -k}b^k$$

That helps a lot in me understanding, thank you!

 

[PeroK]

That helps a lot in me understanding, thank you!

It's more compact that way. Note that the technique of splitting up the sum into two sums, re-indexing one (or both) of them, before re-combining them is a useful idea to remember.