cu n∈N
[n+n+1+2√n(n+1)]=n+n+1+[ 2√n(n+1)]
2n=2√n²≤2√n(n+1)<2√(n+1)(n+1)=2(n+1)=2n+2
cum intre n 2n si 2n+2 exsta inca un nr inttreg, 2n+1,
trebuie sa verificam si daca
2√n(n+1)<2n+1
intr-adevar , prin ridicare la patrat , observam ca
4n²+4n<4n²+4n+1
deci
[2√n(n+1) ]=2n
atunci n+n+1+[ 2√n(n+1)]=n+n+1+2n=4n+1
b)
cu n∈N*
[ (√2+ √(2n²+1))²]=
[2+2n²+1 +2√4n²+2)] =2n²+3 +[√(16n²+8)];
dar
4n=√(16n²)≤√(16n²+8)<√(16n²+8n+1)=√(4n+1)²=4n+1
itr-adevar
16n²+8<16n²+8n+1
pt ca 7<8n, ∀n≥1
deci
[√(16n²+8)]=4n
atunci
2n²+3 +[√(16n²+8)];=2n²+4n+3 ptn∈N*
pt n=0,
[ (√2+ √(2n²+1))²]=5
