tolong y bantu aku,?. terimakasih
Matematika
liska19
Pertanyaan
tolong y bantu aku,?. terimakasih
1 Jawaban
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1. Jawaban newwiguna
4.a) 1 + 3 + 5 + 7 + 9 + ... + 99 = ...?
suku pertama (a) = 1
beda (b) = 3 - 1 = 2
Un = a+(n-1)b
99 = 1 + (n-1)2
99 = 1 + 2n - 2
99 = 2n - 1
2n = 99 + 1
2n = 100
n = 50
Sn = n/2(a+Un)
= 50/2 (1 + 99)
= 25 x 100
= 2.500
1 + 3 + 5 + 7 + 9 + ... + 99 = 2.500
4.b) 1 - 2 + 3 - 4 + 5 - 6 + ... - 100
(1 + 3 + 5 + 7 + ... +99) - (2 + 4 + 6 + 8 + ... +100)
cari jumlah barisan pertama :
1 + 3 + 5 + 7 + 9 + ... + 99 = ...?
suku pertama (a) = 1
beda (b) = 3 - 1 = 2
Un = a+(n-1)b
99 = 1 + (n-1)2
99 = 1 + 2n - 2
99 = 2n - 1
2n = 99 + 1
2n = 100
n = 50
Sn = n/2(a+Un)
= 50/2 (1 + 99)
= 25 x 100
= 2.500
Cari jumlah barisan kedua :
2 + 4 + 6 + 8 + 10 + ... + 100 = ?
a = 2
b = 2
Un = 2 + 2n - 2
= 2n
100 = 2n
n = 100/2
n = 50
S50 = 50/2(2+100)
= 25 x 102
= 2.550
Cari selisihnya = 2.500 - 2.550 = -50
4.c) a = -100
b = 1
Un = -100 + (n-1)1
= -100 + n - 1
50 = n - 101
n = 151
S151 = 151/2 (-100 +50)
= 151/2 x (-50)
= -3.775