How do you expand (2x−3)5 using Pascal’s Triangle?

1 Answer
Surya K.
Jan 26, 2018

32x5−240x4+720x3−1080x2+810x−243

Explanation:

The binomial theorem states, (x+a)n=n∑k=0(n!k!(n−k)!)xn-kak

Here, n=5,x=2x,a=−3

(2x−3)5=5∑k=0(5!k!(5−k)!)(2x)5-k(−3)k

When k=0,

(5!0!(5−0)!)(2x)5-0(−3)0

(1201⋅120)⋅32x5⋅1

=32x5

When k=1,

(5!1!(5−1)!)(2x)5-1(−3)1

1201⋅24⋅16x4⋅(−3)

=−240x4

When k=2,

(5!2!(5−2)!)(2x)5-2(−3)2

1202⋅6⋅8x3⋅9

=720x3

When k=3,

(5!3!(5−3)!)(2x)5-3(−3)3

1206⋅2⋅4x2⋅(−27)

=−1080x2

When k=4,

(5!4!(5−4)!)(2x)5-4(−3)4

12024⋅1⋅2x⋅81

=810x

When k=5,

(5!5!(5−5)!)(2x)5-5(−3)5

120120⋅1⋅1⋅(−243)

=−243

Add each individual answer up. The answer is 32x5−240x4+720x3−1080x2+810x−243.

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