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inductive proof of binomial theorem

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inductive proof of binomial theorem

We prove the theorem for a ring. We do not assume a unit for the ring. We do not need commutativity of the ring, but only that aaa and bbb commute.

When n=1n1n=1, the result is clear.

For the inductive step, assume it holds for mmm. Then for n=m+1nm1n=m+1,

(a+b)m+1superscriptabm1\displaystyle(a+b)^{{m+1}} =\displaystyle= (a+b)⁢(a+b)mabsuperscriptabm\displaystyle(a+b)(a+b)^{m}
=\displaystyle= (a+b)⁢(am+bm+∑k=1m-1(mk)⁢am-k⁢bk)⁢ by the inductive hypothesisabsuperscriptamsuperscriptbmsuperscriptsubscriptk1m1binomialmksuperscriptamksuperscriptbk by the inductive hypothesis\displaystyle(a+b)(a^{m}+b^{m}+\sum_{{k=1}}^{{m-1}}\binom{m}{k}a^{{m-k}}b^{k})% \text{ by the inductive hypothesis}
=\displaystyle= am+1+bm+1+a⁢bm+b⁢am+∑k=1m-1(mk)⁢am-k+1⁢bk+∑k=1m-1(mk)⁢am-k⁢bk+1superscriptam1superscriptbm1asuperscriptbmbsuperscriptamsuperscriptsubscriptk1m1binomialmksuperscriptamk1superscriptbksuperscriptsubscriptk1m1binomialmksuperscriptamksuperscriptbk1\displaystyle a^{{m+1}}+b^{{m+1}}+ab^{m}+ba^{m}+\sum_{{k=1}}^{{m-1}}\binom{m}{% k}a^{{m-k+1}}b^{k}+\sum_{{k=1}}^{{m-1}}\binom{m}{k}a^{{m-k}}b^{{k+1}}
=\displaystyle= am+1+bm+1+∑k=1m(mk)⁢am-k+1⁢bk+∑k=0m-1(mk)⁢am-k⁢bk+1⁢ by combining termssuperscriptam1superscriptbm1superscriptsubscriptk1mbinomialmksuperscriptamk1superscriptbksuperscriptsubscriptk0m1binomialmksuperscriptamksuperscriptbk1 by combining terms\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\binom{m}{k}a^{{m-k+1}}b^{k}% +\sum_{{k=0}}^{{m-1}}\binom{m}{k}a^{{m-k}}b^{{k+1}}\text{ by combining terms}
=\displaystyle= am+1+bm+1+∑k=1m(mk)⁢am-k+1⁢bk+∑j=1m(mj-1)⁢am+1-j⁢bj⁢ let j=k+1 in second sumsuperscriptam1superscriptbm1superscriptsubscriptk1mbinomialmksuperscriptamk1superscriptbksuperscriptsubscriptj1mbinomialmj1superscriptam1jsuperscriptbj let j=k+1 in second sum\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\binom{m}{k}a^{{m-k+1}}b^{k}% +\sum_{{j=1}}^{m}\binom{m}{j-1}a^{{m+1-j}}b^{j}\text{ let j=k+1 in second sum}
=\displaystyle= am+1+bm+1+∑k=1m[(mk)+(mk-1)]⁢am+1-k⁢bk⁢ by combining the sumssuperscriptam1superscriptbm1superscriptsubscriptk1mbinomialmkbinomialmk1superscriptam1ksuperscriptbk by combining the sums\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\left[\binom{m}{k}+\binom{m}% {k-1}\right]a^{{m+1-k}}b^{k}\text{ by combining the sums}
=\displaystyle= am+1+bm+1+∑k=1m(m+1k)⁢am+1-k⁢bk⁢ from Pascal’s rulesuperscriptam1superscriptbm1superscriptsubscriptk1mbinomialm1ksuperscriptam1ksuperscriptbk from Pascal’s rule\displaystyle a^{{m+1}}+b^{{m+1}}+\sum_{{k=1}}^{m}\binom{m+1}{k}a^{{m+1-k}}b^{% k}\text{ from Pascal's rule}

as desired.

Keywords: 
number theory
Major Section: 
Reference
Type of Math Object: 
Proof
Parent: 
binomial theorem

Mathematics Subject Classification

05A10 Factorials, binomial coefficients, combinatorial functions

Comments

Submitted by russ42897 on Sun, 11/09/2003 - 23:18

I wanted some hints on this problem. Let f be a bounded function on interval I, prove that
sup({-f(x): x in I}) = -inf({f(x): x in I}).

Submitted by vernondalhart on Mon, 11/10/2003 - 05:31

We had a problem similar to this in our honors calculus class last year; The way I found best was to start by contradiction, that they are not equal.

That is, if f is a function f: I -> A, then there is some element a in A such that a < -inf({f(x) : x in I}) with a an upper bound for {-f(x): x in I}.

What then, can you say about -a and its relation to {f(x) : x in I}

Well, at least something like that.

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Owner: Mathprof
Added: 2001-10-18 - 19:47
Author(s): Mathprof

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(v21) by Mathprof 2013-03-22
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