a) Aratati ca \( n+1<\sqrt{n(n+3)}<n+2 \) , \( \forall n\in \mathb{N} \) , \( n>1 \)
b)Calculati suma \( S=[\sqrt{2\cdot 5}]+[\sqrt{3\cdot 6}]+[\sqrt{4\cdot 7}+...+[\sqrt{100\cdot 103}] \) , unde [...] reprezinta partea intreaga a numarului .
c)Determinati \( n\in \mathb{N} \) astfel incat \( [(\sqrt{n}+\sqrt{n+3})^2]=2001 \)
Indicatie : a)Ridicare la patrat
b)\( n+1\le \sqrt{n(n+3)}<n+2 \)
c)Dezvoltarea parantezelor
Concursul interjudetean de matematica ''Marian Tarina'' p.II
Concursul interjudetean de matematica ''Marian Tarina'' p.II
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Andi Brojbeanu
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a) Ridicam la patrat egalitatea si obtinem:
(1)\( (n+1)^2<n(n+3) \)\( ; n^2+2n+1<n^2+3n \); \( 1<n \), adevarat din ipoteza.
(2)\( n(n+3)<(n+2)^2 \)\( ; n^2+3n<n^2+4n+4 \)\( ; 0<n+4 \), adevarat.
b) Din a obtinem ca \( [\sqrt{n(n+3})]=n+1 \), si atunci S devine:
\( 2+1+3+1+.......+100+1=(1+2+3+......+100)-1+99\cdot \)\( 1 \)\( =50\cdot \)\( 101-1+99 \)\( =5148 \).
c)\( [(\sqrt{n}+\sqrt{n+3})^2] \)\( =[(\sqrt{n})^2+(\sqrt{n+3})^2+2\sqrt{n(n+3)}] \)\( =[n+n+3]+[2\sqrt{n(n+3)}] \)\( =2n+3+2n+2+4n+5 \);\( 4n+5=2001 \)\( ;4n=1996 \)\( ; n=499 \).
(1)\( (n+1)^2<n(n+3) \)\( ; n^2+2n+1<n^2+3n \); \( 1<n \), adevarat din ipoteza.
(2)\( n(n+3)<(n+2)^2 \)\( ; n^2+3n<n^2+4n+4 \)\( ; 0<n+4 \), adevarat.
b) Din a obtinem ca \( [\sqrt{n(n+3})]=n+1 \), si atunci S devine:
\( 2+1+3+1+.......+100+1=(1+2+3+......+100)-1+99\cdot \)\( 1 \)\( =50\cdot \)\( 101-1+99 \)\( =5148 \).
c)\( [(\sqrt{n}+\sqrt{n+3})^2] \)\( =[(\sqrt{n})^2+(\sqrt{n+3})^2+2\sqrt{n(n+3)}] \)\( =[n+n+3]+[2\sqrt{n(n+3)}] \)\( =2n+3+2n+2+4n+5 \);\( 4n+5=2001 \)\( ;4n=1996 \)\( ; n=499 \).