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Brazilian Mathematical Olympiad Difficulty 4.9 AIME Prove it Brazil
a a a b b b c c c d d d a d ≠ b c ad \neq bc a d = b c 1 ( a x + b ) ( c x + d ) \frac{1}{(ax+b)(cx+d)} ( a x + b ) ( c x + d ) 1 r a x + b + s c x + d \frac{r}{ax+b} + \frac{s}{cx+d} a x + b r + c x + d s 1 1 ⋅ 4 + 1 4 ⋅ 7 + 1 7 ⋅ 10 + ⋯ + 1 2998 ⋅ 3001 \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{2998 \cdot 3001} 1 ⋅ 4 1 + 4 ⋅ 7 1 + 7 ⋅ 10 1 + ⋯ + 2998 ⋅ 3001 1
Solution a a x + b − c c x + d = a d − b c ( a x + b ) ( c x + d ) , so we can take r = a a d − b c , s = − c a d − b c . \frac{a}{ax+b} - \frac{c}{cx+d} = \frac{ad-bc}{(ax+b)(cx+d)}, \text{ so we can take } r = \frac{a}{ad-bc}, s = -\frac{c}{ad-bc}. a x + b a − c x + d c = ( a x + b ) ( c x + d ) a d − b c , so we can take r = a d − b c a , s = − a d − b c c .
∑ k = 0 999 1 ( 3 k + 1 ) ( 3 k + 4 ) = 1 3 ∑ k = 0 999 ( 1 3 k + 1 − 1 3 k + 4 ) = 1 3 ( 1 − 1 3001 ) = 1000 3001 \sum_{k=0}^{999} \frac{1}{(3k+1)(3k+4)} = \frac{1}{3} \sum_{k=0}^{999} \left( \frac{1}{3k+1} - \frac{1}{3k+4} \right) = \frac{1}{3} \left( 1 - \frac{1}{3001} \right) = \frac{1000}{3001} k = 0 ∑ 999 ( 3 k + 1 ) ( 3 k + 4 ) 1 = 3 1 k = 0 ∑ 999 ( 3 k + 1 1 − 3 k + 4 1 ) = 3 1 ( 1 − 3001 1 ) = 3001 1000
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