Page 52 - Buku Siap OSN Matematika SMP 2015(1)

Aljabar

2010
n = = 1005
2

n
Sn =  2a  n  1   b
2

1005

4
=  2   1005 1 3    
2
 
1005
=   6 1004    4
2
1005
=   6 4016  
2

1005
=   4022  = –1005  2011
2

Sehingga:
2
2
2
2
2
2
2
1 – 2 + 3 – 4 + 5 – ... – x 2010 + 2011 = –1005  2011 + 2011  2011
1
= 2011  ( 2011 – 1005)
= 2011  1006
= 2023066
2
2
2
2
2
2
2
Jadi, nilai dari: 1 – 2 + 3 – 4 + 5 – ... – 2010 + 2011 = 2023066
2
2
2
2
2
2
2
2
2
8. Jika nilai 100B = 100 + 99 – 98 – 97 + 96 + 95 – 94 – 93 +  + 4 + 3 2
2
2
– 2 – 1 , maka nilai B adalah ...
Jawab:
2
2
2
2
2
2
2
2
2
2
2
100B = 100 + 99 – 98 – 97 + 96 + 95 – 94 – 93 +  + 4 + 3 – 2 – 1 2
2
2
2
2
2
2
2
2
2
2
100B = (100 – 98 ) + (99 – 97 ) + (96 – 94 ) + (95 – 93 ) +  + (4 – 2 )
2
2
+ (3 – 1 )
100B = (100 – 98)  (100 + 98) + (99 – 97)  (99 + 97) + (96 – 94)  (96 + 94)
+ (95 – 93)  (95 + 93) +  + (4 – 2)  (4 + 2) + (3 – 1)  (3 + 1)
100B = 2  198 + 2  196 + 2  190 + 2  188 +  + 2  6 + 2  4
100B = 2  (198 + 196 + 190 + 188 +  + 6 + 4)
Siap OSN Matematika SMP 2015 43