Binomial Theorem (Binomial Expansion)

pull into
main question
reach

pull into

There is first a widely known conclusion:

\[(a+b)^2=a^2+2ab+b^2 \]

So, how does one seek\((a+b)^3\) And? Hand-calculated, as follows:

\[\begin{aligned} (a+b)^3 &= (a+b)\times(a+b)^2\\ &=(a+b)\times(a^2+2ab+b^2)\\ &=[a\times(a^2+2ab+b^2)]+[b\times(a^2+2ab+b^2)]\\ &=(a^3+2a^2b+ab^2)+(a^2b+2ab^2+b^3)\\ &=a^3+3a^2b+3ab^2+b^3 \end{aligned} \]

main question

Try to be experimental and look for patterns, and then you'll find the

\[(a+b)^k=\sum\limits_{i=0}^{k}\binom{k}{i}a^ib^{k-i} \]

included among these\(\binom{k}{i}\)expressed in\(k\) Take any one of the following\(i\) The number of programs that are equivalent to the number of\(C_{k}^{i}\)。
Proof:
First, if we don't combine like terms, then we get the number of terms\(2^k\). This can be proved this way: suppose it is known\((a+b)^k\) In the case of non-consolidation\(n\) item, then

\[\begin{aligned} (a+b)^{k+1} &= (a+b)(a+b)^k\\ &=a(a+b)^k+b(a+b)^k \end{aligned}\]

The number of terms in both the left and right equations is\(n\)and is\(2n\)。
after that\(a\) cap (a poem)\(b\) The sum of the number of times the\(k\), this conclusion is obvious. When we multiply, each\((a+b)\) I'll pick one of them.\(a\) maybe\(b\)out. So hanging in front of that combination of numbers also proves out: take to the\(a^ib^{k-i}\) The case is equivalent to the case where there is a length of\(k\) The 01 sequence of 1's with the number of 1's\(i\), the problem of finding the number of different scenarios is clearly\(C_i^k\). Similarly:

exist\(k\) It's on in one of the lights.\(i\) Shut it down.\(k-i\) number of programs to solve the problem of finding the number of programs.
Find the number of paths from (1,1) to (i,j) in the grid map, which can only go up and to the right.

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However, the Binomial Theorem can be extended to much more than that. Check this out:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
………………
Above.\((a+b)^k\) Table of coefficients. Does it feel like a bit of déjà vu?

Yes, that's it, the Yang Hui Triangle!
Next, we are going to prove at the binomial level that the Yang-Fai triangle can be used to find the coefficients of a binomial after expansion. It is not a bad idea to start by assuming that the Yang-Fai triangle can be used to find the coefficients after the expansion, so the Yang-Fai triangle of the first\(i\) banker's anti-fraud numeral\(j\)The columns can then be represented as\(C_i^j\), in combination with the recursive formula for finding Yang Hui's triangle\(a_{i,j}=a_{i-1,j-1}+a_{i-1,j}\)There is this style:

\[C_i^j=C_{i-1}^{j-1}+C_{i-1}^{j} \]

Then push the right form as follows:

\[\begin{aligned} C_{i-1}^{j-1}+C_{i-1}^{j}&=\tfrac{(i-1)!}{(i-j)!(j-1)!}+\tfrac{(i-1)!}{(i-j-1)!j!}\\&=\tfrac{1}{j!}(\tfrac{(i-1)!j}{(i-j)!}+\tfrac{(i-1)!}{(i-j-1)!})\\&=\tfrac{1}{j!(i-j)!}[(i-1)!j+(i-1)!(i-j)]\\&=\tfrac{(a-1)!}{j!(i-j)!}(b+a-b)\\&=\tfrac{(a-1)!a}{j!(i-j)!}(b+a-b)\\&=\tfrac{a!}{j!(i-j)!}\\&=C_i^j \end{aligned}\]

So Yang Hui's recursion of the triangle to find the coefficients of the binomial expansion is correct. This also gives us another idea for recursively finding combinatorial numbers, which would be useful in some topics, but I don't think anyone would normally choose to\(O(n)\) Solve for combinatorial numbers, I guess.