Tentukan hasil dari (tanpa menghitung satu persatu) a. 1 + 3 + 5 + 7 + 9 + ... + 99

Jawaban Ayo Kita Berlatih 1.2 Halaman 20 - 21 Matematika Kelas 7. Latihan 1.2 Halaman 17 - 20 Matematika Kelas 7.

B. Soal Uraian

3. Tentukan hasil dari (tanpa menghitung satu persatu)

a. 1 + 3 + 5 + 7 + 9 + ... + 99

b. 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + ... − 100

c. −100 − 99 − 98 − … − 2 − 1 − 0 + 1 + 2 + ... + 48 + 49 + 50

Penyelesaian:

A)

Sn = 1+3+5+7+9...+99

Sn = U1 + U2 + U3... + Un

a = U1 = 1

b = U2 - U1 → 2

Un = a + (n-1)b

99 = 1 + (n-1)2

99 = 1 + 2n - 2

99 = 2n - 1

2n = 100

n = 100/2

n = 50

Sn = (n/2)(a + Un)

S50 = (50/2)(a + U50)

S50 = 25(1+99)

S50 = 25(100)

S50 = 2500

Kesimpulan :

" 1+3+5..+99 = 2500 "

B)

Sn = 1-2+3-4+5-6...-100

Sn = (1+3+5..+99) - (2+4+6..+100)

Sn = Sn.Deret 1 - Sn.Deret 2

#Deret 1

Sn = 1+3+5..+99

Sn = U1+U2+U3..+Un

a = U1 → 1

b = U2 - U1 → 2

Un = a + (n-1)b

99 = 1 + (n-1)2

n = 100/2

n = 50

Sn = (n/2)(a+Un)

S50 = (50/2)(1+99)

S50 = 25(100)

S50 = 2500

Sn.Deret 1 = 2500

#Deret 2

Sn = 2+4+6...+100

a = 2

b = 2

Un = a + (n-1)b

100 = 2 +(n-1)2

n = 100/2

n = 50

Sn = (n/2)(a+Un)

S50 = (50/2)(2+100)

S50 = 25(102)

S50 = 2550

Sn.Deret 2 = 2550

Sn = Sn.Deret 1 - Sn.Deret 2

Sn = 2500 - 2550

Sn = -50

Kesimpulan :

" 1-2+3-4...+99-100 = -50 "

C) Sn = -100-99...-1-0+1... +50

nilai positif hanya pada interval 1-50, sedangkan negatif pada interbal 1-100, maka : ↓

= -100-99..-1-0+1..+50

= -100-99..-51 + (-50...-1-0+1...+50)

= -100-99...-51

= - (100+99..+51)

= - Sn

a = 100

b = -1

Un = a + (n-1)b

51 = 100 +(n-1)-1

-49 = -n +1

n = 49+1

n = 50

-Sn = (50/2)(100+51)

-Sn = 25(151)

-Sn = 3775

Sn = 3775/-1

Sn = -3775

Kesimpulan :

" -100-99...+49+50 = -3775 "

Jawaban Pertanyaan dari Buku MTK Kelas 7 Semester 1. Ayo Kita Berlatih 1.2 Halaman 20, latihan 1.2 Halaman 17.

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