Question

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

Answer 1

A)

#Deret Aritmatika

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)

#Deret Aritmatika

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)

#Deret Aritmatika

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 "