Pentru un numar natural \( n \), notam cu \( u(n) \) cel mai mare numar prim mai mic sau egal cu \( n \) si \( v(n) \) cel mai mic numar prim mai mare decat \( n \). Sa se arate ca:
\( \frac{1}{u(2)v(2)}+\frac{1}{u(3)v(3)}+\frac{1}{u(4)v(4)}+...+\frac{1}{u(2010)v(2010)}=\frac{1}{2}-\frac{1}{2011} \)
Numere prime
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Marius Mainea
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Daca \( p \) si \( q \) sunt numere prime consecutive , notam \( A_p=\{x\in\mathb{N} | p\le x<q} \) . Se oberva ca \( A_p \) are \( q-p \) elemente si pentru orice termen \( x \) al sau avem \( u(x)=p \) si \( v(x)=q \) . Cu alte cuvinte , termenul \( \frac{1}{p\cdot q} \) apare in suma de \( q-p \) ori . Suma devine :
\( \frac{3-2}{2\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{2011-2003}{2003\cdot2011}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2011}=\frac{1}{2}-\frac{1}{2011} \)
\( \frac{3-2}{2\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{2011-2003}{2003\cdot2011}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2011}=\frac{1}{2}-\frac{1}{2011} \)
. A snake that slithers on the ground can only dream of flying through the air.