Rumus Binomial Newton adalah sebagai berikut.
(a+b)n=i=0∑n(ni)a(n−i)bi
Dengan
a=2pb=p23
Maka,
0!(12−0)!12!(2p)12(p23)0=====0!(12−0)!12!(2p)12⋅10!(12−0)!12!(2p)1212!12!⋅(2p)121⋅212p124.096p12
1!(12−1)!12!(2p)11(p23)1=======1!⋅ +4mu11!12!(2p)11(p23)11!12⋅11!⋅211p11⋅(p23)p212⋅ +4mu3⋅ +4mu211p11211⋅ +4mu36p11−2211⋅3692.048⋅ +4mu36p973.728p9
2!(12−2)!12!(2p)10(p23)2========2!⋅ +4mu10!12!2!⋅ +4mu10!12⋅11⋅10!⋅210⋅p10⋅(p23)22!12⋅ +4mu11⋅210⋅p10⋅(p432)2!12⋅ +4mu11⋅p4132⋅ +4mu32⋅ +4mu210p102!210⋅ +4mu32⋅ +4mu132p10−42⋅11.024⋅9⋅132p621.216.512p6606.256p6
3!(12−3)!12!(2p)9(p23)3=3!⋅ +4mu9!12!⋅29p9⋅p633p9=3!⋅9!12⋅ +4mu11⋅ +4mu10⋅9!⋅29p9⋅p633p9=3!12⋅ +4mu11⋅ +4mu10⋅⋅29p9⋅p633p9=3!p61.320⋅ +4mu33⋅ +4mu29p9=3!29⋅ +4mu33⋅ +4mu1.320p9−6=3!29⋅ +4mu33⋅ +4mu1.320p3=3⋅2⋅1512⋅27⋅1.320p2=618.247.680p3=3.041.280p3
4!(12−4)!12!(2p)8(p23)4=========4!(12−4)!12!⋅28p8⋅(p2)4344!⋅ +4mu8!12!⋅28p8⋅p8344!⋅8!12⋅ +4mu11⋅ +4mu10⋅ +4mu9⋅8!⋅28p8⋅p8344!12⋅ +4mu11⋅ +4mu10⋅ +4mu9⋅28p8⋅p8344!11.880⋅28⋅p834p84⋅3⋅2⋅111.880⋅28⋅p834p82411.880⋅256⋅81495⋅256⋅8110.264.320
Jadi, suku yang bebas p pada penjabaran (2p+p23)12 adalah
.
Oleh karena itu, jawaban yang tepat adalah C.