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Question

What is the coefficient for the 5th term of (2x-y)⁸

Answer 1

1

11

121

1331

14641

15101051

1615201561

172135352171

1-8-28-56-70-56-28-8-1

Pascal's Triangle, 9th row 5th term is 70

in the Triange each number is the sum of 2 digits just above it

70 is the coefficent of the 5th term of (2x+y)^8 BUT more likely the coefficient you want is 1120, see below

(2x)^8 + 8(2x)^7(y) + 28(2x)^6(y^2) + 56(2x)^5(y^3) + 70(2x)^4(y^4) + ....

5th term is 70(2x)^4(y^4)

= 70(16)x^4(y^4)

= 1120x^4y^4 = the 5th term

or to find the coefficient

calculate 8C4 = 8!/4!4! = 8x7x6x5/4x3x2 = 8x7x6x5/4x6 = 70= same as 5th term of 9th row of Pascal's Triangle

first term without the coefficient is (2x)^8, 2nd is (2x)^7y 3rd is (2x)^6y^2 4th is (2x)^5y^3 5th is (2x)^4y^4

x exponents decrease by 1 each term, y exponents increase by 1 each term. Their exponents sum to 8

with coefficient 70, 5th term is 70(2x)^4y^4 = 1120x^4(y^4)

nth term of (a+b)^n = nC(n-r) = n!/r!k! x^r-1(y^(n-r+1) = 70(2x)^4(y^4) = 1120x^4(y^4)

IF you really have nothing else to do other than very tedious multiplication multiply out

(2x+y)(2x+y)(2x+y)(2x+y)(2x+y)(2x+y)(2x+y)(2x+y) = (2x)^8 + 8(2x)^7(y) + 28(2x)^6)y^2 + 56(2x)^5)y^3 + 70(2x)^4)y^4, + ....

where r = 5, n= 8, k= n-r+1 = 8-5+1=4