The binomial theorem states, (x+a)^n=sum_"k=0"^n((n!)/(k!(n-k)!))x^"n-k"a^k(x+a)n=n∑k=0(n!k!(n−k)!)xn-kak
Here, n=5, x=2x,a=-3n=5,x=2x,a=−3
(2x-3)^5=sum_"k=0"^5((5!)/(k!(5-k)!))(2x)^"5-k"(-3)^k(2x−3)5=5∑k=0(5!k!(5−k)!)(2x)5-k(−3)k
When k=0k=0,
((5!)/(0!(5-0)!))(2x)^"5-0"(-3)^0(5!0!(5−0)!)(2x)5-0(−3)0
(120/(1*120))*32x^5*1(1201⋅120)⋅32x5⋅1
=32x^5=32x5
When k=1k=1,
((5!)/(1!(5-1)!))(2x)^"5-1"(-3)^1(5!1!(5−1)!)(2x)5-1(−3)1
120/(1*24)*16x^4*(-3)1201⋅24⋅16x4⋅(−3)
=-240x^4=−240x4
When k=2k=2,
((5!)/(2!(5-2)!))(2x)^"5-2"(-3)^2(5!2!(5−2)!)(2x)5-2(−3)2
120/(2*6)*8x^3*91202⋅6⋅8x3⋅9
=720x^3=720x3
When k=3k=3,
((5!)/(3!(5-3)!))(2x)^"5-3"(-3)^3(5!3!(5−3)!)(2x)5-3(−3)3
120/(6*2)*4x^2*(-27)1206⋅2⋅4x2⋅(−27)
=-1080x^2=−1080x2
When k=4k=4,
((5!)/(4!(5-4)!))(2x)^"5-4"(-3)^4(5!4!(5−4)!)(2x)5-4(−3)4
120/(24*1)*2x*8112024⋅1⋅2x⋅81
=810x=810x
When k=5k=5,
((5!)/(5!(5-5)!))(2x)^"5-5"(-3)^5(5!5!(5−5)!)(2x)5-5(−3)5
120/(120*1)*1*(-243)120120⋅1⋅1⋅(−243)
=-243=−243
Add each individual answer up. The answer is 32x^5-240x^4+720x^3-1080x^2+810x-24332x5−240x4+720x3−1080x2+810x−243.
