What is the binomial theorem?

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What is the binomial theorem?

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Answer 1 · 2016-02-12T19:55:00

It is a method one may use to expand a binomial expression raised to a positive integer power as follows :

${(x + y)}^{n} = n \sum r = 0^{n} C_{r} x^{n - r} y^{r}$ $(x+y)^n=sum_(r=0)^n""^nC_rx^(n-r)y^r$

The combination notation used is defined as follows :

$^{n} C_{r} = \frac{n!}{(n - r)! r!}$ $""^nC_r=(n!)/((n-r)!r!)$

Example :

Expand ${(2 x + 3 y)}^{5}$ $(2x+3y)^5$ .

This is a binomial (2 terms) raised to an integer power, so the binomial theorem is valid and may be used as follows :

${(2 x + 3 y)}^{5} = 5 \sum r = 0^{5} C_{r} {(2 x)}^{5 - r} {(3 y)}^{r}$ $(2x+3y)^5=sum_(r=0)^5 ""^5C_r(2x)^(5-r)(3y)^r$

$=^{5} C_{0} {(2 x)}^{5 - 0} {(3 y)}^{0} +^{5} C_{1} {(2 x)}^{5 - 1} {(3 y)}^{1} +^{5} C_{2} {(2 x)}^{5 - 2} {(3 y)}^{2} +^{5} C_{3} {(2 x)}^{5 - 3} {(3 y)}^{3} +^{5} C_{4} {(2 x)}^{5 - 4} {(3 y)}^{4} +^{5} C_{5} {(2 x)}^{5 - 5} {(3 y)}^{5}$ $=""^5C_0(2x)^(5-0)(3y)^0+""^5C_1(2x)^(5-1)(3y)^1+""^5C_2(2x)^(5-2)(3y)^2+""^5C_3(2x)^(5-3)(3y)^3+""^5C_4(2x)^(5-4)(3y)^4+""^5C_5(2x)^(5-5)(3y)^5$

$= (1) 32 x^{5} (1) + (5) (16 x^{4}) (3 y) + (10) (8 x^{3}) (9 y^{2}) + (10) (4 x^{2}) (27 y^{3}) + (5) (2 x) (81 y^{4}) + (1) (1) (243 y^{5})$ $=(1)32x^5(1)+(5)(16x^4)(3y)+(10)(8x^3)(9y^2)+(10)(4x^2)(27y^3)+(5)(2x)(81y^4)+(1)(1)(243y^5)$

$= 32 x^{5} + 240 x^{4} y + 720 x^{3} y^{2} + 1080 x^{2} y^{3} + 810 x y^{4} + 243 y^{5}$ $=32x^5+240x^4y+720x^3y^2+1080x^2y^3+810xy^4+243y^5$ .

Answer 2 · 2016-02-16T21:09:23

There is a simpler way of expanding a binomial that uses the binomial theorem but takes a more intuitive approach.

Instead of doing $_{n} C_{r}$ $""_n"C"_r$ in front of each term, we can use the coefficients in the $n + 1$ $n+1$ row of Pascal's triangle.

In the case of ${(2 x + 3 y)}^{5}$ $(2x+3y)^5$ , the $_{n} C_{r}$ $""_n"C"_r$ series will be identical to the $6$ $6$ th row of Pascal's triangle:

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The row we want is

$1, 5, 10, 10, 5, 1$ $1,5,10,10,5,1$

In order to deal with exponents, know that the exponent on the first term will start at $5$ $5$ and work its way down to $0$ $0$ , and the second term will start with an exponent of $0$ $0$ and work its way up to $5$ $5$ .

If there is a negative term they will alternate positive, negative, positive, negative, etc.

For ${(2 x + 3 y)}^{5}$ $(2x+3y)^5$ , we get

$1 {(2 x)}^{5} {(3 y)}^{0} + 5 {(2 x)}^{4} {(3 y)}^{1} + 10 {(2 x)}^{3} {(3 y)}^{2} + 10 {(2 x)}^{2} {(3 y)}^{3} + 5 {(2 x)}^{1} {(3 y)}^{4} + 1 {(2 x)}^{0} {(3 y)}^{5}$ $1(2x)^5(3y)^0+5(2x)^4(3y)^1+10(2x)^3(3y)^2+10(2x)^2(3y)^3+5(2x)^1(3y)^4+1(2x)^0(3y)^5$

Note that anything to the $0$ $0$ power is $1$ $1$ .

Simplified, this gives us

$32 x^{5} + 240 x^{4} y + 720 x^{3} y^{2} + 1080 x^{2} y^{3} + 810 x^{4} + 243 y^{5}$ $32x^5+240x^4y+720x^3y^2+1080x^2y^3+810x^4+243y^5$

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What is the binomial theorem?

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