Jawaban Ulangan I

LOGIKA & INDUKSI MATEMATIKA

11 𝑀𝐼𝑃𝐴 1, 2, 3, 4, 5

11 MIPA 1

1 E 6 E 11 C 16 E 21 B2 B 7 E 12 D 17 B 22 B3 B 8 D 13 D 18 B 23 E4 C 9 C 14 B 19 D 24 D5 E 10 E 15 D 20 C 25 D

𝑃+: 3- .+ − 1 = 8 → 𝑏𝑒𝑛𝑎𝑟, ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 8

1) 𝐵𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑃B: 3-B − 1 ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 8

𝑃C: 3-C − 1

11 MIPA 1

𝑃CD+: 3-(𝒌D𝟏) − 1 = 3-CD- − 1 = 3- . 3-C − 1

= 9 . 3-C − 1 = 8 . 3-C + 1 . 3-C − 1

J

ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 8

J

ℎ𝑎𝑏𝑖𝑠 8

2) 𝐽𝑖𝑘𝑎 𝑛 ∈ 𝑐𝑎𝑐𝑎ℎ, 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎11 MIPA 1

𝑃B: 1 + 2 + 2- + 2O + . . . . . . + 2B = 2BD+ − 1

𝑃+: 1 = 2+D+ −1 → 𝑏𝑒𝑛𝑎𝑟

𝑃C: 1 + 2 + 2- + 2O + . . . . . . + 2C = 2CD+ − 1

𝑃𝒌D𝟏: 1 + 2 + 2- + 2O + . . . . . . + 2C + 2𝒌D𝟏

= 2CD+ − 1 + 2𝒌D𝟏

= 2 . 2CD+ − 1

= 2CD- − 1 = 2𝒌D𝟏D+ − 1

𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖

3) 𝐽𝑖𝑘𝑎 𝑎 ∈ 𝑏𝑖𝑙. 𝑔𝑎𝑛𝑗𝑖𝑙, 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎11 MIPA 1

𝑎- + 3𝑎 + 5 𝑎𝑑𝑎𝑙𝑎ℎ 𝑏𝑖𝑙. 𝑔𝑎𝑛𝑗𝑖𝑙 𝑗𝑢𝑔𝑎

𝑔𝑎𝑛𝑗𝑖𝑙 → 𝑎 = 2𝑘 + 1

𝑎- + 3𝑎 + 5 = 2𝑘 + 1 - + 3 2𝑘 + 1 + 5

= 4𝑘- + 4𝑘 + 1 + 6𝑘 + 3 + 5

= 4𝑘- + 10𝑘 + 9

= 2 2𝑘- + 5𝑘 + 4 + 1 = 2𝑚 + 1

𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 𝑔𝑎𝑛𝑗𝑖𝑙

4) 𝐽𝑖𝑘𝑎 𝑛 ∈ 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖 & 𝑛 ≥ 2

𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 5B + 9 < 6B → 𝑃B

𝑃-: 5- + 9 < 6- → 34 < 36 𝑏𝑒𝑛𝑎𝑟

𝑃C: 5C + 9 < 6C

𝟔 5C + 9 < 𝟔 . 6C →𝑘𝑎𝑙𝑖 𝟔 → (𝟓 + 𝟏) 5C + 9 < 6CD+

𝑡𝑢𝑗𝑢𝑎𝑛: 𝟓𝒌D𝟏 + 𝟗 < 𝟔𝒌D𝟏

5CD+ + 45 + 5C + 9 < 6CD+ → 5CD+ + 9 + 5C + 45 < 6CD+

𝑘𝑎𝑟𝑒𝑛𝑎 5C + 45 > 0 𝑚𝑎𝑘𝑎 5CD+ + 9 + 0 < 6CD+

𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖

11 MIPA 1

1 E 6 E 11 C 16 C 21 B2 B 7 E 12 D 17 B 22 B3 B 8 D 13 D 18 B 23 E4 C 9 C 14 E 19 D 24 D5 E 10 E 15 D 20 C 25 D

11 MIPA 2

11 MIPA 2

𝑃+: 5- .+ + 3 − 1 = 25 + 3 − 1 = 27 → 𝑏𝑒𝑛𝑎𝑟, ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 9

1) 𝐵𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑃B: 5-B + 3𝑛 − 1 ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 9

𝑃C: 5-C + 3𝑘 − 1

𝑃𝒌D𝟏: 5-(𝒌D𝟏) + 3 𝒌 + 𝟏 − 1 = 5-CD- + 3𝑘 + 3 − 1

= 25 . 5-C +3𝑘 + 3 − 1

= 24 . 5-C + 3 + 5-C + 3𝑘 − 1J

ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 9= 3 8 . 5-C + 1ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 3

11 MIPA 2

𝑃B: 4007B − 1 ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 2003

2) 𝐽𝑖𝑘𝑎 𝑛 ∈ 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖, 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎

𝑃+: 4007+ − 1 = 4006 → 𝑏𝑒𝑛𝑎𝑟, ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 2003

𝑃C: 4007C − 1

𝑃𝒌D𝟏: 4007𝒌D𝟏 − 1 = 4007 . 4007C − 1

= 4006 . 4007C + 4007C − 1J

ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 2003Jℎ𝑎𝑏𝑖𝑠 2003

11 MIPA 2

𝑎- + 2𝑎 + 3 𝑎𝑑𝑎𝑙𝑎ℎ 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑔𝑒𝑛𝑎𝑝3) 𝐽𝑖𝑘𝑎 𝑎 ∈ 𝑏𝑖𝑙. 𝑔𝑎𝑛𝑗𝑖𝑙, 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎

𝑔𝑎𝑛𝑗𝑖𝑙 → 𝑎 = 2𝑘 + 1

𝑎- + 2𝑎 + 3 = 2𝑘 + 1 - + 2 2𝑘 + 1 + 3

= 4𝑘- + 4𝑘 + 1 + 4𝑘 + 2 + 3

= 4𝑘- + 8𝑘 + 6

= 2 2𝑘- + 4𝑘 + 3 = 2𝑚 𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 𝑔𝑒𝑛𝑎𝑝

11 MIPA 2

𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 3B > 2𝑛 + 14) 𝐽𝑖𝑘𝑎 𝑛 ∈ 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖 & 𝑛 ≥ 2 → 𝑃B

𝑃-: 3- > 2 . 2 + 1 → 𝑏𝑒𝑛𝑎𝑟

𝑃C: 3C > 2𝑘 + 1 𝑡𝑢𝑗𝑢𝑎𝑛: 𝟑𝒌D𝟏 > 𝟐 𝒌 + 𝟏 + 𝟏

𝑘𝑎𝑙𝑖 𝟑 → 3 . 3C > 3(2𝑘 + 1)

3CD+ > 6𝑘 + 3

3CD+ > 2𝑘 + 3 + 4𝑘 𝑘𝑎𝑟𝑒𝑛𝑎 4𝑘 > 0

3CD+ > 2𝑘 + 3 + 0 → 3CD+ > 2 𝑘 + 1 + 1𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖

11 MIPA 3

1 E 6 E 11 C 16 C 21 B

2 B 7 E 12 D 17 B 22 B

3 B 8 D 13 D 18 B 23 E

4 C 9 C 14 E 19 D 24 D

5 E 10 E 15 D 20 C 25 D

11 MIPA 31) 𝐵𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑃B: 2-BD+ + 1 ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 3

𝑃+: 2- .+D+ + 1 = 8 + 1 → 𝑏𝑒𝑛𝑎𝑟, ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 3

𝑃C: 2-CD+ + 1

𝑃CD+: 2-(CD+)D+ + 1 = 2-CD-D+ + 1

= 4 . 2-CD+ + 1

= 3 . 2-CD+ + 2-CD++ 1J

ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 3Jℎ𝑎𝑏𝑖𝑠 3

11 MIPA 32) 𝐽𝑖𝑘𝑎 𝑛 ∈ 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛

𝑃B: 41B − 14B 𝑎𝑑𝑎𝑙𝑎ℎ 𝑘𝑒𝑙𝑖𝑝𝑎𝑡𝑎𝑛 27

𝑃+: 41+ − 14+ = 27 → 𝑏𝑒𝑛𝑎𝑟, 𝑘𝑒𝑙𝑖𝑝𝑎𝑡𝑎𝑛 27

𝑃C: 41C − 14C

𝑃𝒌D𝟏: 41𝒌D𝟏 − 14𝒌D𝟏 = 41 . 41C − 14 . 14C

= 27 . 41C + 14 . 41C − 14 . 14C

= 27 . 41C + 14 41C − 14CJ

ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 27J

ℎ𝑎𝑏𝑖𝑠 27

11 MIPA 33) 𝐵𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎 ℎ𝑎𝑠𝑖𝑙 𝑘𝑎𝑙𝑖 𝑑𝑢𝑎 𝑏𝑖𝑙. 𝑔𝑎𝑛𝑗𝑖𝑙

𝑎𝑑𝑎𝑙𝑎ℎ 𝑏𝑖𝑙. 𝑔𝑎𝑛𝑗𝑖𝑙 𝑗𝑢𝑔𝑎

𝑚𝑖𝑠𝑎𝑙 𝑘𝑒𝑑𝑢𝑎 𝑏𝑖𝑙. 𝑖𝑡𝑢: 𝑎 = 2𝑚 + 1 & 𝑏 = 2𝑛 + 1

𝑎 . 𝑏 = (2𝑚 + 1)(2𝑛 + 1)

= 4𝑚𝑛 + 2𝑚 + 2𝑛 + 1

= 2 2𝑚𝑛 +𝑚 + 𝑛 + 1

= 2𝑘 + 1

𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 𝑔𝑎𝑛𝑗𝑖𝑙

11 MIPA 34) 𝐽𝑖𝑘𝑎 𝑛 𝜖 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖 & 𝑛 ≥ 5 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 4𝑛 < 2B

𝑃a: 4 . 5 < 2a → 𝑏𝑒𝑛𝑎𝑟

𝑃C: 4𝑘 < 2C

𝑃CD+: 4(𝑘 + 1) < 2CD+

2CD+ ≥ 2 . 2C ≥ 2C + 2C

𝑘 ≥ 5

2C ≥ 2a 𝑡𝑒𝑛𝑡𝑢 𝑑𝑜𝑛𝑔

2C ≥ 32 > 4

2C > 4

> 2C + 4𝑘 > 4 + 4𝑘

11 MIPA 4

1 E 6 E 11 C 16 E 21 B

2 B 7 E 12 D 17 B 22 B

3 B 8 D 13 D 18 B 23 E

4 C 9 C 14 B 19 D 24 D

5 E 10 E 15 D 20 C 25 D

11 MIPA 4

1) Buktikan 𝑃B: 5B − 1 habis dibagi 4

𝑃+: 5+ − 1 = 4 → 𝑏𝑒𝑛𝑎𝑟, ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 4

𝑃C: 5C − 1

𝑃CD+: 𝟓𝒌D𝟏 − 𝟏= 5 . 5C − 1= 4 . 5C + 𝟓𝒌 − 𝟏

↓ ↓ℎ𝑎𝑏𝑖𝑠 4 ℎ𝑎𝑏𝑖𝑠 4

2) Jika n ∈ bil asli maka buktikan bahwa𝑃B: 2 + 4 + 6 + . . . . + 2𝑛 = 𝑛- + 𝑛

𝑃+ = 2 = 1- + 1 = 1 → 𝑏𝑒𝑛𝑎𝑟

𝑃C: 2 + 4 + 6 + . . . . + 2𝑘 = 𝒌𝟐 + 𝒌

𝑃𝒌D𝟏: 2 + 4 + 6 + . . . . + 2𝑘 + 2 𝒌 + 𝟏

= 𝒌𝟐 + 𝒌 + 2𝑘 + 2 = 𝑘- + 2𝑘 + 1 + 𝑘 + 1= 𝒌 + 𝟏 - + 𝒌 + 𝟏 𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖

11 MIPA 4

3) Jika 𝑎 adalah bilangan ganjil maka buktikan 𝑎Oadalah bilangan ganjil juga.

𝑎 = 2𝑘 + 1

𝑎O = 2𝑘 + 1 O = 𝟏 . 2𝑘 O + 𝟑 2𝑘 -. 1 + 𝟑 2𝑘 . 1- + 𝟏

= 8𝑘O + 12𝑘- + 6𝑘 + 1

= 2 4𝑘O + 6𝑘- + 3𝑘 + 1

= 2𝑚 + 1𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 𝑎O 𝑏𝑖𝑙𝑎𝑛𝑔𝑎𝑛 𝑔𝑎𝑛𝑗𝑖𝑙

11 MIPA 4

4) Jika 𝑛 ∈ 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖 & 𝑛 ≥ 3 buktikan 𝑛 + 1 - < 2𝑛-

𝑃O: 3 + 1 - < 2 . 3- → 4- < 18 → 𝑏𝑒𝑛𝑎𝑟

𝑃C: 𝑘 + 1 - < 2𝑘- 𝑡𝑢𝑗𝑢𝑎𝑛: 𝒌 + 𝟏 + 𝟏 𝟐< 𝟐 𝒌 + 𝟏 𝟐

11 MIPA 4

𝑘- + 2𝑘 + 1 < 2𝑘-

𝑘- + 2𝑘 + 1 + 2𝑘 + 3 < 2𝑘- + 2𝑘 + 3

𝑘- + 2𝑘 + 1 + 2𝑘 + 3 < 2𝑘- + 2𝑘 + 3 + 2𝑘 − 1

𝑘- + 4𝑘 + 4 < 2 𝑘- + 2𝑘 + 1

𝑘 + 1 + 1 - < 2 𝑘 + 1 - 𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 = 𝑡𝑢𝑗𝑢𝑎𝑛

11 MIPA 5

1 E 6 E 11 E 16 C 21 B

2 B 7 E 12 D 17 B 22 B

3 B 8 D 13 D 18 B 23 E

4 C 9 C 14 E 19 D 24 D

5 E 10 E 15 D 20 C 25 D

1) 𝐵𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑃B: 3-B + 2-BD- ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 5

𝑃+: 3- .+ + 2- .+D- = 9 + 16 = 25 → 𝑏𝑒𝑛𝑎𝑟, ℎ𝑎𝑏𝑖𝑠 𝑑𝑖𝑏𝑎𝑔𝑖 5

𝑃C: 3-C+ 2-CD-

𝑃𝒌D𝟏: 3-(𝒌D𝟏)+ 2-(𝒌D𝟏)D-

= 3-CD- + 2-CD-D- = 9 . 3-C + 4 . 2-CD-

= 5 . 3-C + 𝟒 . 3-C + 𝟒 . 2-CD- = 5 . 3-C + 𝟒 3-C + 2-CD-

11 MIPA 5

J

ℎ𝑎𝑏𝑖𝑠 𝑏𝑎𝑔𝑖 5

J

ℎ𝑎𝑏𝑖𝑠 5

11 MIPA 52) 𝐽𝑖𝑘𝑎 𝑛 ∈ 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛

𝑃B: 𝑛- + 2𝑛 𝑎𝑑𝑎𝑙𝑎ℎ 𝑘𝑒𝑙𝑖𝑝𝑎𝑡𝑎𝑛 3

𝑃+: 1- + 2 . 1 = 3 → 𝑏𝑒𝑛𝑎𝑟, 𝑘𝑒𝑙𝑖𝑝𝑎𝑡𝑎𝑛 3

𝑃C: 𝑘- + 2𝑘

𝑃CD+: (𝑘 + 1)- + 2 𝑘 + 1 = 𝑘- + 2𝑘 + 1 + 2𝑘 + 2

= 𝑘- +2𝑘 + 2𝑘 + 3J

𝑘𝑒𝑙𝑖𝑝𝑎𝑡𝑎𝑛 3

J

𝑏𝑢𝑘𝑎𝑛 𝑘𝑒𝑙𝑖𝑝𝑎𝑡𝑎𝑛 3𝑡𝑎𝑘 𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 ‼

11 MIPA 53) 𝐽𝑖𝑘𝑎 𝑎 ∈ 𝑏𝑖𝑙. 𝑔𝑎𝑛𝑗𝑖𝑙, 𝑏𝑢𝑘𝑡𝑖𝑘𝑎𝑛 𝑏𝑎ℎ𝑤𝑎

𝑎- + 𝑎 + 3 𝑎𝑑𝑎𝑙𝑎ℎ 𝑏𝑖𝑙. 𝑔𝑎𝑛𝑗𝑖𝑙 𝑗𝑢𝑔𝑎

𝑔𝑎𝑛𝑗𝑖𝑙 → 𝑎 = 2𝑘 + 1

𝑎- + 𝑎 + 1 = 2𝑘 + 1 - + 2𝑘 + 1 + 1

= 4𝑘- + 4𝑘 + 1 + 2𝑘 + 2

= 4𝑘- + 6𝑘 + 2 + 1

= 2 2𝑘- + 3𝑘 + 1 + 1 = 2𝑚 + 1

𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 𝑔𝑎𝑛𝑗𝑖𝑙

11 MIPA 5

4. Jika 𝑛 ∈ 𝑏𝑖𝑙. 𝑎𝑠𝑙𝑖 & 𝑛 ≥ 2 buktikan 4𝑛 < 2B → 𝑃B

𝑃-: 4 . 2 < 2- → 8 < 4 → 𝑠𝑎𝑙𝑎ℎ

𝑡𝑎𝑘 𝑡𝑒𝑟𝑏𝑢𝑘𝑡𝑖 ‼

Jadwal POST TEST

11 MIPA 1 Selasa, 24 Sept11 MIPA 2 Senin, 23 Sept11 MIPA 3 Senin, 23 Sept11 MIPA 4 Senin, 23 Sept11 MIPA 5 Selasa, 24 Sept

Perhitungan

Pre Test Post Test Ambil60 40 6060 90 9080 40 8080 90 90