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Suku yang bebas p pada penjabaran ( 2 p + p 2 3 ​ ) 12 adalah ...

Suku yang bebas  pada penjabaran  adalah ...

  1. begin mathsize 14px style 10.156.320 end style 

  2. begin mathsize 14px style 10.240.320 end style 

  3. begin mathsize 14px style 10.264.320 end style 

  4. begin mathsize 14px style 10.371.240 end style 

  5. begin mathsize 14px style 10.394.120 end style 

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G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

jawaban yang tepat adalah C.

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Pembahasan

Rumus Binomial Newton adalah sebagai berikut. ( a + b ) n = i = 0 ∑ n ​ ( n i ​ ) a ( n − i ) b i Dengan a = 2 p b = p 2 3 ​ Maka, i = 0 0 ! ( 12 − 0 ) ! 12 ! ​ ( 2 p ) 12 ( p 2 3 ​ ) 0 ​ = = = = = ​ 0 ! ( 12 − 0 ) ! 12 ! ​ ( 2 p ) 12 ⋅ 1 0 ! ( 12 − 0 ) ! 12 ! ​ ( 2 p ) 12 12 ! 12 ! ​ ⋅ ( 2 p ) 12 1 ⋅ 2 12 p 12 4.096 p 12 ​ i = 1 1 ! ( 12 − 1 ) ! 12 ! ​ ( 2 p ) 11 ( p 2 3 ​ ) 1 ​ = = = = = = = ​ 1 ! ⋅ + 4 m u 11 ! 12 ! ​ ( 2 p ) 11 ( p 2 3 ​ ) 11 ! 12 ⋅ 11 ! ​ ⋅ 2 11 p 11 ⋅ ( p 2 3 ​ ) p 2 12 ⋅ + 4 m u 3 ⋅ + 4 m u 2 11 p 11 ​ 2 11 ⋅ + 4 m u 36 p 11 − 2 2 11 ⋅ 3 6 9 2.048 ⋅ + 4 m u 36 p 9 73.728 p 9 ​ i = 2 2 ! ( 12 − 2 ) ! 12 ! ​ ( 2 p ) 10 ( p 2 3 ​ ) 2 ​ = = = = = = = = ​ 2 ! ⋅ + 4 m u 10 ! 12 ! ​ 2 ! ⋅ + 4 m u 10 ! 12 ⋅ 11 ⋅ 10 ! ​ ⋅ 2 10 ⋅ p 10 ⋅ ( p 2 3 ​ ) 2 2 ! 12 ⋅ + 4 m u 11 ​ ⋅ 2 10 ⋅ p 10 ⋅ ( p 4 3 2 ​ ) 2 ! 12 ⋅ + 4 m u 11 ​ ⋅ p 4 132 ⋅ + 4 m u 3 2 ⋅ + 4 m u 2 10 p 10 ​ 2 ! 2 10 ⋅ + 4 m u 3 2 ⋅ + 4 m u 132 p 10 − 4 ​ 2 ⋅ 1 1.024 ⋅ 9 ⋅ 132 p 6 ​ 2 1.216.512 p 6 ​ 606.256 p 6 ​ i = 3 3 ! ( 12 − 3 ) ! 12 ! ​ ( 2 p ) 9 ( p 2 3 ​ ) 3 = 3 ! ⋅ + 4 m u 9 ! 12 ! ​ ⋅ 2 9 p 9 ⋅ p 6 3 3 ​ p 9 = 3 ! ⋅ 9 ! 12 ⋅ + 4 m u 11 ⋅ + 4 m u 10 ⋅ 9 ! ​ ⋅ 2 9 p 9 ⋅ p 6 3 3 ​ p 9 = 3 ! 12 ⋅ + 4 m u 11 ⋅ + 4 m u 10 ​ ⋅ ⋅ 2 9 p 9 ⋅ p 6 3 3 ​ p 9 = 3 ! p 6 1.320 ⋅ + 4 m u 3 3 ⋅ + 4 m u 2 9 p 9 ​ = 3 ! 2 9 ⋅ + 4 m u 3 3 ⋅ + 4 m u 1.320 p 9 − 6 ​ = 3 ! 2 9 ⋅ + 4 m u 3 3 ⋅ + 4 m u 1.320 p 3 ​ = 3 ⋅ 2 ⋅ 1 512 ⋅ 27 ⋅ 1.320 p 2 ​ = 6 18.247.680 p 3 ​ = 3.041.280 p 3 i = 4 4 ! ( 12 − 4 ) ! 12 ! ​ ( 2 p ) 8 ( p 2 3 ​ ) 4 ​ = = = = = = = = = ​ 4 ! ( 12 − 4 ) ! 12 ! ​ ⋅ 2 8 p 8 ⋅ ( p 2 ) 4 3 4 ​ 4 ! ⋅ + 4 m u 8 ! 12 ! ​ ⋅ 2 8 p 8 ⋅ p 8 3 4 ​ 4 ! ⋅ 8 ! 12 ⋅ + 4 m u 11 ⋅ + 4 m u 10 ⋅ + 4 m u 9 ⋅ 8 ! ​ ⋅ 2 8 p 8 ⋅ p 8 3 4 ​ 4 ! 12 ⋅ + 4 m u 11 ⋅ + 4 m u 10 ⋅ + 4 m u 9 ​ ⋅ 2 8 p 8 ⋅ p 8 3 4 ​ 4 ! 11.880 ​ ⋅ 2 8 ⋅ p 8 3 4 ​ p 8 4 ⋅ 3 ⋅ 2 ⋅ 1 11.880 ​ ⋅ 2 8 ⋅ p 8 ​ 3 4 ​ p 8 ​ 24 11.880 ​ ⋅ 256 ⋅ 81 495 ⋅ 256 ⋅ 81 10.264.320 ​ Jadi, suku yang bebas p pada penjabaran ( 2 p + p 2 3 ​ ) 12 adalah . Oleh karena itu, jawaban yang tepat adalah C.

Rumus Binomial Newton adalah sebagai berikut.

Dengan

Maka,

 

 

Jadi, suku yang bebas  pada penjabaran  adalah begin mathsize 14px style 10.264.320 end style .

 

Oleh karena itu, jawaban yang tepat adalah C.

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