Binomial expansion

Question

What are the first three terms in the expansion (2 - 10x)^5?

Answer 1

5C5(2^5) = 1(32) = 32 = the 1st term

5C4(2^4)(-10x) = 5(16)(-10x) = -800x = the 2nd term

5C3(2^3)(-10x)^2 = 10(8)(100x^2) = 8,000x^2 = 3rd term

first 3 terms are 32, -800x, 8000x^2

5C2(2^2)(-10x)^3 = 10(4)(-1000x^3) = -40000x^3 = 4th term

5C1(2)(-10x)^4 = 5th term

5C0(-10x)^5 = last term

nCr(2)^n(-10x)^r = the nth term

nCr = n!/r!(n-r)!

5C3 = 5!3!2! = 5x4x3x2x1/3x2x1x2x1 = 10

Pascal's Triangle also gives the coefficients = nCr

1 for (a+b)^0 = 1

1,1 for (a+b)^1= a+b

1,2,1 for (a+b)^2 = a^2 +2ab + b^2

1,3,3,1 for (a+b)^3 = a^3 +3a^2b + 3ab^2 + b^3

1,4,6,4,1 for (a+b)^4= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

1,5,10,10,5,1 for (a+b)^5= a^5 +5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

substitute 2 for a and -10x

to get the terms of the binomial expansion of (2-10x)^5 = (2^5) + 5(2^4)(-10x) + 10(2^3)(-10x)^2 + ...