1+2+3+4+5-6+...+1999=
Matematika
Riefqy1
Pertanyaan
1+2+3+4+5-6+...+1999=
2 Jawaban
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1. Jawaban Grace99
n = 1999
a = 1
b = 2 - 1 = 1
Sn = n/2(2a + (n - 1)b)
S1999 = 1999/2(2 . 1 + (1999 - 1) . 1)
S1999 = 1999/2(2 + 1998)
S1999 = 1999/2(2000)
S1999 = 1999 . 1000
S1999 = 1999000
1 + 2 + 3 + 4 + 5 + 6 + ... + 1999 = 1999000 -
2. Jawaban ErikCatosLawijaya
Mapel : Matematika
Kelas : IX SMP
Bab : Barisan dan Deret
Pembahasan :
1 + 2 + 3 + 4 + ... + 1999
Sn = n/2(2a + (n - 1)b)
S1999 = 1999/2(2.1 + (1998))
S1999 = 1999/2(2000)
S1999 = 1.999.000