Ppg: Kiprah M1 Kb2 Matematika Kombinatorik

M1 KB2 KOMBINATORIK
1. Ekspansikan dengan teorema Binomial Newton:
(a) (1 + 𝑥) −1
(1 + 𝑥) −1 = ∑𝐶(−1, 𝑘 )𝑥 𝑘 ∞ 𝑘=0 = ∑(−1) 𝑘𝐶(1 + 𝑘 − 1, 𝑘)𝑥 𝑘 ∞ 𝑘=0
= 1 − 𝑥 + 𝑥 2 − 𝑥 3 + 𝑥 4 − ⋯

(b) (1 + 𝑥) −2
(1 + 𝑥) −2 = ∑𝐶(−2, 𝑘 )𝑥 𝑘 ∞ 𝑘=0 = ∑(−1) 𝑘𝐶(2 + 𝑘 − 1, 𝑘)𝑥 𝑘 ∞ 𝑘=0
= 1 − 2𝑥 + 3𝑥 2 − 4𝑥 3 + 5𝑥 4 − ⋯

(c) (1 + 𝑥) −3
(1 + 𝑥) −3 = ∑𝐶(−3, 𝑘 )𝑥 𝑘 ∞ 𝑘=0 = ∑(−1) 𝑘𝐶(3 + 𝑘 − 1, 𝑘)𝑥 𝑘 ∞ 𝑘=0
= 1 − 3𝑥 + 6𝑥 2 − 10𝑥 3 + 15𝑥 4 − ⋯

2. Koefesien 𝑥 101𝑦 49 dari (2𝑥 − 3𝑦) 150
(𝑥 + 𝑦) 𝑛 = ∑𝐶(𝑛, 𝑘 )𝑥 𝑛−𝑘𝑦 𝑘
𝑛 𝑘=0 𝑥 101𝑦 49
𝑛 − 𝑘 = 101 dan 𝑘 = 49
𝐶(150,49 )(2) 101(−3) 49 = 150! 49! .101! 2 101(−3) 49

3. 𝐴 = {4. 𝑎, 3. 𝑏, 2. 𝑐}
𝐵 = {2. 𝑎, 3. 𝑏, 4. 𝑐}
𝐴 ∪ 𝐵 = {4. 𝑎, 3. 𝑏, 4. 𝑐}
𝐴 ∩ 𝐵 = {2. 𝑎, 3. 𝑏, 2. 𝑐}
𝐴 − 𝐵 = {2. 𝑎}

4. 𝑎𝑛 = 2𝑎𝑛−1 + 𝑎𝑛−2 − 2𝑎𝑛−3
𝑎𝑛 = 2𝑎𝑛−1 + 𝑎𝑛−2 − 2𝑎𝑛−3
𝑎𝑛 − 2𝑎𝑛−1 − 𝑎𝑛−2 + 2𝑎𝑛−3 = 0
𝑟 3 − 2𝑟 2 − 𝑟 1 + 2𝑟 0 = 0
𝑟 3 − 2𝑟 2 − 𝑟 + 2 = 0
(𝑟 + 1)(𝑟 − 1)(𝑟 − 2) = 0
𝑟1 = −1, 𝑟2 = 1, 𝑟3 = 2

Solusi relasi rekursif
𝑎𝑛 = 𝑐1. 𝑟1 𝑛 + 𝑐2. 𝑟2 𝑛 + 𝑐3. 𝑟3 𝑛
𝑎𝑛 = 𝑐1. (−1) 𝑛 + 𝑐2. 1 𝑛 + 𝑐3. 2 𝑛

𝑎0 = 9 = 𝑐1 + 𝑐2 + 𝑐3 persamaan 1
𝑎1 = 10 = −𝑐1 + 𝑐2 + 2𝑐3 persamaan 2
𝑎2 = 32 = 𝑐1 + 𝑐2 + 4𝑐3 persamaan 3
Persamaan 3 dan 1
𝑐1 + 𝑐2 + 4𝑐3 = 32
𝑐1 + 𝑐2 + 𝑐3 = 9
_________________-
3𝑐3 = 23
𝑐3 = 23/3

𝑐1 + 𝑐2 + 23/3 = 9
3𝑐1 + 3𝑐2 + 23 = 27
3𝑐1 + 3𝑐2 = 4 persamaan 4
−𝑐1 + 𝑐2 + 2. 23/3 = 10
−3𝑐1 + 3𝑐2 + 46 = 30
−3𝑐1 + 3𝑐2 = −16 persamaan 5

Persamaan 4 dan 5
3𝑐1 + 3𝑐2 = 4
−3𝑐1 + 3𝑐2 = −16
__________________ +
 6𝑐2 = −12
𝑐2 = −2

3𝑐1 + 3𝑐2 = 4
3𝑐1 + 3(−2) = 4
3𝑐1 − 6 = 4 3𝑐1 = 10
𝑐1 = 10/3

𝑎𝑛 = 10 3 . (−1) 𝑛 − 2. 1 𝑛 + 23 3 . 2 𝑛 = 10 3 . (−1) 𝑛 + 23 3 . 2 𝑛 − 2
Makara solusi relasi rekursif dari 𝑎𝑛 = 2𝑎𝑛−1 + 𝑎𝑛−2 − 2𝑎𝑛−3 adalah
𝑎𝑛 = 10 3 . (−1) 𝑛 + 23 3 . 2 𝑛 − 2

5. 𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2
𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2
𝑓𝑛 − 𝑓𝑛−1 − 𝑓𝑛−2 = 0
𝑟 2 − 𝑟 1 − 𝑟 0 = 0
𝑟 2 − 𝑟 1 − 1 = 0
𝑟 2 − 𝑟 1 = 1
 𝑟 2 − 𝑟 1 + 1 4 = 1 + 1 4
(𝑟 − 1/2 )2 = 5/4
𝑟 − 1/2 = ±√ 5/4
𝑟 − 1 2 = ± 1/2 √5 𝑟 = 1 / ± 1/2 √5 𝑟1 = 1/2 + 1/2 √5 𝑎𝑡𝑎𝑢 𝑟2 = 1/2 − 1/2 √5

Solusi relasi untuk dua akar kembar
𝑓𝑛 = 𝑐1. 𝑟1 𝑛 + 𝑐2. 𝑟2 𝑛
𝑓𝑛 = 𝑐1. ( 1/2 + 1/2 √5) 𝑛 + 𝑐2. ( 1/2 − 1/2 √5) 𝑛
𝑓0 = 0 = 𝑐1. ( 1/2 + 1/2 √5) 0 + 𝑐2. ( 1/2 − 1/2 √5) 0 = 𝑐1 + 𝑐2
𝑓1 = 1 = 𝑐1. ( 1/2 + 1/2 √5) 1 + 𝑐2. ( 1/2 − 1/2 √5) 1 = ( 1/2 + 1/2 √5) 𝑐1 + ( 1/2 − 1/2 √5) 𝑐2
( 1/2 + 1/2 √5) 𝑐1 + ( 1/2 − 1/2 √5) 𝑐2 = 1
( 1/2 + 1/2 √5) 𝑐1 + ( 1/2 + 1/2 √5) 𝑐2 = 0
_______________________________ -
−√5𝑐2 = 1
𝑐2 = 1/−√5 = − √5/5
𝑐1 + 𝑐2 = 0
𝑐1 − √5/5 = 0
𝑐1 = √5/5

𝑓𝑛 = 𝑐1. ( 1/2 + 1/2 √5) 𝑛 + 𝑐2. ( 1/2 − 1/2 √5) 𝑛
𝑓𝑛 = √5/5 . ( 1/2 + 1/2 √5) 𝑛 + (− √5/5 ) . ( 1/2 − 1/2 √5) 𝑛
𝑓𝑛 = ( √5/10 + 5/10) 𝑛 + ( −√5/10 + 5/10) 𝑛
Makara relasi solusi rekursif dari 𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2 ialah 𝑓𝑛 = ( √5/10 + 5/10) 𝑛 + ( −√5/10 + 5/10) 𝑛

6. Fungsi pembangkit dari 0, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0, ...
𝑎𝑛 = 2(0,1,1,1,1,1,1,1,0,0,0,0)
𝐺(𝑥) = 2(1 + 𝑥 1 + 𝑥 2 + 𝑥 3 + 𝑥 4 + 𝑥 5 + 𝑥 6 + 𝑥 7 ) = 2 ( 1 − 𝑥 7 1 − 𝑥 )

7. Barisan dari fungsi pembangkit
(a) (1 + 3𝑥) −1
(1 + 3𝑥) −1 = 1/1 + 3𝑥 = 1/1 − (−3𝑥)
1/1 − (−3𝑥) = ∑(−3) 𝑘𝑥 𝑘 ∞ 𝑘=0 = (−3) 0𝑥 0 + (−3) 1𝑥 1 + (−3) 2𝑥 2 + (−3) 3𝑥 3 + ⋯
 = 1, −3𝑥 + 9𝑥 2 − 27𝑥 3 + ⋯
Sehingga barisannya ialah {1, -3, 9, -27, ...}

(b) (1 − 𝑥) −2
(1 − 𝑥) −2 = 1 (1 − 𝑥) 2
1 (1 − 𝑥) 2 = ∑(𝑘 + 1)𝑥 𝑘 ∞ 𝑘=0 = 1𝑥 0 + 2𝑥 1 + 3𝑥 2 + 4𝑥 3 + ⋯
= 1 + 2𝑥 + 3𝑥 2 + 4𝑥 3 + ⋯
Sehingga barisannya ialah {1, 2, 3, 4, ...}

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