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 Grace99n = 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
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			  	2. Jawaban ErikCatosLawijayaMapel : 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