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Algebra Level 2

( 1 + 2 − 3 ) + ( 4 + 5 − 6 ) + . . . + ( 2998 + 2999 − 3000 ) 500 = ? \large\frac{(1+2-3)+(4+5-6)+...+(2998+2999-3000)}{500} = \ ?


The answer is 2997.

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Jade Mijares by Jade Mijares , 30, Philippines

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1 solution

Jade Mijares
Mar 2, 2015

( 1 + 2 − 3 ) = 0 (1+2-3) = 0 , ( 4 + 5 − 6 ) = 3 (4+5-6) = 3 , ( 7 + 8 − 9 ) = 6 (7+8-9) = 6 , ( 10 + 11 − 12 ) = 9 (10+11-12) = 9 , so their sums is a arithmetic sequence. Then, n ( 2 a 1 + ( n − 1 ) d 2 ) = 1000 ( 2 ( 0 ) + ( 1000 − 1 ) 3 2 ) = 1498500 n\bigg(\frac{2a_1 + (n - 1)d}{2}\bigg) = 1000\bigg(\frac{2(0) + (1000 - 1)3}{2}\bigg)= 1498500 Then 1498500 500 = 2997 \frac{1498500}{500} = \boxed {2997}

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