Exercise 62 Page 714

We want to find the second term of the expansion of the given expression. (4x-5)^(10) ⇔ ( 4x+( -5))^(10)To do so, we will use the Binomial Theorem to write the expansion in sigma notation. (a+b)^n=∑_(k=0)^n n!/k!(n-k)!a^(n-k)b^k For our expression, we have that a= 4x, b= - 5, and n= 10. Since k starts at 0, for the second term we have k=1. Remember that we do not have to find the sum — we only need to find one of the terms. Therefore, we will only consider the argument of the summation notation instead of the entire sum.

n!/k!(n-k)!a^(n-k)b^k

SubstituteValues

Substitute values

10!/1!( 10-1)!( 4x)^(10-1)( - 5)^()darkorange1

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Simplify

SubTerms

Subtract terms

10!/1!(9!)(4x)^9(- 5)^1

1!=1

10!/9!(4x)^9(- 5)^1

n!=n* (n-1)!

10(9!)/9!(4x)^9(- 5)^1

CancelCommonFac

Cancel out common factors

10(4x)^9(- 5)^1

PowProdII

(a b)^m=a^m b^m

10(4)^9x^9(- 5)^1

ExponentOne

a^1=a

10(4)^9x^9(- 5)

CalcPow

Calculate power

10(262 144)x^9(- 5)

MultPosNeg

a(- b)=- a * b

- 13 107 200x^9