Lee asked in Science & MathematicsMathematics · 2 months ago

# How do I get the Newtonian binomial expanded form of this equation?

(a1+a2+a3+....an)^m

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• nbsale
Lv 6
2 months ago

Each term involves the variables a1 ... an, raised to a power. The power may be 0.

The powers add up to m. Call the p1 ... pn.

p1 + ... + pn = m

The coefficient is n! / (p1! X p2! X ... X pn!)

E.g., (a+b+c)^5 = a^5 + 5 a^4 b + 5 a^4 c + 10 a^3 b^2 + 20 a^3 b c + 10 a^3 c^2 + 10 a^2 b^3 + 30 a^2 b^2 c + 30 a^2 b c^2 + 10 a^2 c^3 + 5 a b^4 + 20 a b^3 c + 30 a b^2 c^2 + 20 a b c^3 + 5 a c^4 + b^5 + 5 b^4 c + 10 b^3 c^2 + 10 b^2 c^3 + 5 b c^4 + c^5

I like to think of it this way.

(a+b+c)^5 = (a+b+c)(a+b+c)(a+b+c)(a+b+c)(a+b+c)

When you multiply that out, you have to pick a variable from each (a+b+c).  You have to make m choices, which is why the powers (exponents) have to add up to m.

If you pick a every time, you get a^5, and there is only 1 combination that gets you a^5. So you have the term a^5 or 1a^5.

If you pick a 4 times, and b once, you get a^4(b), and there are 5 ways to do that (being the 5 possible choices for the b). That's why the answer above has the term 5 a^4 b.

The number of ways to get, say, a^2 b^2 c is

5! / 2!2!1! = 120/4 = 30. (You have the total factorial, 5!, divided by each exponent factorial. So you have the term

30 a^2 b^2 c

All this is a simple, and pretty cool, IMHO, expansion of the binomial theorem. It's very handy to know when you have n possible outcomes for m trials.

• 2 months ago

binomial (a+b)ⁿ

polinomial (a⑴+a⑵+...+a⒩)ⁿ