Explain the steps of binomial expansion, I just learned about binomial theorem but i'm having trouble understanding the expansion process. Please help!

Answer 1

in the general expansion of (a+b)^n

1. If n is a whole number:
(a+b)^1 = a+b
(a+b)^2 = a^2 + 2ab + b^2
(a+b)^3 = a^3 + 3a^2 b + 3 a b^2 + b^3
(a+b)^4 = a^4 + 4a^3 b + 6a^2 b^2 + 4 a b^3 + b^4
...
notice that the coefficients are the terms of rows in Pascal's Triangle, you must
absolutely familiarize yourself with that

in general for whole numbers of n
(a+b)^n = C(n,n)a^n + C(n,n-1) a^(n-1) b + C(n,n-2) a^(n-2) b^2 + .... +C(n,0) b^n
e.g.
(2x - 3y)^4 = C(4,4)(2x)^4 (-3y)^0 + C(4,3)(2x)^3 (-3y)^1 + C(4,2)(2x)^2 (-3y)^2 + C(4,1)(2x)^1 (-3y)^3 + C(4,0)(2x)^0 (-3y)^4
= 1(16x^4)(1) + 4(8x^3)(-3y) + 6(4x^2)(9y^2) + 4(2x)(-27y^3) + (1)(1)(81y^4)
= 16x^4 - 96x^3 y + 216x^2 y^2 - 35x y^3 + 81y^4

2. if n is not a whole number, but could be any real number, then
(a+b)^r = a^r + r(a^(r-1) b + r(r-1)/2! (a^(r-2)) (b^2) + r(r-1)(r-2)/3! (a^(r-3))(b^3) + r(r-1)(r-2)(r-3)/4! (a^(r-4))(b^4) + ....
Can you see the pattern ?

e.g. (2x - 3y)^-5
= (2x)^-5 + (-5)(2x)^-6 (-3y) + (-5)(-6)/2! (2x)^-7 (-3y)^2 + (-5)(-6)(-7)/3! (2x)^-8 (-3y)^3 + ... never ends
= (1/32)(1/x^5) + (15/64)(y/x^6) + (135/128)(y^2 / x^7) + (945/256)(y^3/x^8) + ....

In the "olden days" we would use this to find roots of numbers.
e.g. find cube root of 30 to 3 decimals
30^(1/3) < find the closest perfect cube to 30 which is 27
= (27 + 3)^(1/3)
= 27^(1/3) (1 + 3/27)^(1/3) <<< factored out 27^(1/3), which I know
= 3 (1 + 1/9)^(1/3) <---- now expand
= 3[1^(1/3) + (1/3)(1)(1/9) + (1/3)(1/3-1)/2! (1)(1/9)^2 + (1/3)(-2/3)(-5/3)/3! (1) (1/9)^3 + ...]
=3 [ 1 + 1/27 - 1/729 + 5/59049 - .... ]
= 3[ 1 + .03703... - .00137.. + .0000846 ..... ]
= 3[1.03574..]
= 3.10723..
my calculator gave me 3.107232506
Hey, not bad, eh?