1)Suma a oricare \( 2k+1 \) numere consecutive se divide prin \( 2k+1 \) ?
2)Suma a oricare \( 2k \) numere consecutive se divide prin \( 2k \) ?
3)Care este forma patratelor perfecte ? (adica cum sunt numerele impare \( 2k+1 \))
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Humuhumunukunukuapua
La 3 vezi aici : http://mateforum.ro/viewtopic.php?t=2747
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Andi Brojbeanu
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1)Numerele sunt: \( x, x+1, x+2, ......, x+2k-1, x+2k \).
Atunci suma lor va fi egala cu: \( x+x+1+x+2+....+x+2k-1+x+2k=(2k+1)\cdot x+(1+2+3+.....+2k-1+2k)=(2k+1)\cdot x+\frac{2k\cdot (2k+1)}{2}=(2k+1)(x+k)\vdots 2k+1 \).
2)Numerele sunt: \( x, x+1, x+2, ......, x+2k-2, x+2k-1 \).
Atunci suma lor va fi egala cu: \( x+x+1+x+2+....+x+2k-2+x+2k-1=2k\cdot x+(1+2+3+.....+2k-2+2k-1)=2k\cdot x+\frac{(2k-1)2k}{2}=2k(x+\frac{2k-1}{2}) \).
Deoarece \( \frac{2k-1}{2}\notin \mathb{N} \)\( \Rightarrow 2k \) nu divide \( 2k(x+\frac{2k-1}{2}) \).
Atunci suma lor va fi egala cu: \( x+x+1+x+2+....+x+2k-1+x+2k=(2k+1)\cdot x+(1+2+3+.....+2k-1+2k)=(2k+1)\cdot x+\frac{2k\cdot (2k+1)}{2}=(2k+1)(x+k)\vdots 2k+1 \).
2)Numerele sunt: \( x, x+1, x+2, ......, x+2k-2, x+2k-1 \).
Atunci suma lor va fi egala cu: \( x+x+1+x+2+....+x+2k-2+x+2k-1=2k\cdot x+(1+2+3+.....+2k-2+2k-1)=2k\cdot x+\frac{(2k-1)2k}{2}=2k(x+\frac{2k-1}{2}) \).
Deoarece \( \frac{2k-1}{2}\notin \mathb{N} \)\( \Rightarrow 2k \) nu divide \( 2k(x+\frac{2k-1}{2}) \).