Given the binomial (2-5x)^8

Question

(2-5x)⁸

a. Determine the middle term for this expansion. State the answer in the simplest form.

b. If the expansion is written in ascending order of powers, determine the seventh term. (1 mark)

Answer 1

(2-5x)^8

= 2^8 + (8)(2)^7(-5x) + (8!/6!2!)(2)^6(-5x)^2 + (8!/5!3!)(2^5(-5x)^3 + (8!/4!4!)(2^4)(-5x^4) +(8!/3!5!)(2^3))(-5x^5)

+ (8!/2!6!)(2^2)(-5x)^6 + (8)(2)(-^5x)^7 + (-5x)^8

5th term is the middle term = (8!/4!4!)(2^4)(-5x^4) = (8)(7)(6)(5)/(4)(3)(2)](16)(625)x^4 = (70(16)(625)x^5

= 700,000x^4

= (8C4)(16)(625)x^4= 10,000(8C4)x^4

7th term is (8C6)(4)(15625)x^6, = 28(4)(15625)x^6 = 1,750,000x^6

the nth term of the binomial expansion is 8Cn(2)^(9-n)(-5x)^n = [(8!/n!(8-n)!](2^(9-n)(-5x)^n

you can also find the coefficients of the terms from Pascal's Triangle

1

11

121

1331

14641

15101051

1615201561

172135352171

18285670562881

9th row is

1 8 28 56 70 56 28 8 1, which are the coefficients of the 9 terms of (a+b)^8 =a^8 + 8a^7b + 28a^6b^2 + ...

which are the same as

8C8 8C7 8C6 8C5 8C4 8C3 8C2 8C1 8C0

I won't guarantee the arithmetic, but this is the basic idea.

Answer 2

a. Total terms are 8 + 1 = 9. So, fifth term is middle.

T₅ = C(8, 4)·28-⁴(-5x)⁴ = 70·16·625x⁴ = 700000x⁴

b. T₇ = C(8, 6)·2²·(- 5x)⁶ = 28·4·15625x⁶ = 1750000x⁶