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demonstrati ca
a)[tex]\frac{1}{n+1} +\frac{1}{n+2} +...+ \frac{1}{3n+1} \ \textgreater \ 1 , n\geq 1[/tex]
b)[tex]1+\frac{1}{3} +\frac{1}{7} +...+ \frac{1}{2^{n}-1}\leq n[/tex] [tex]n\geq 1[/tex]
c)[tex]\frac{1}{n+1} +\frac{1}{n+2} +...+\frac{1}{2n}\ \textgreater \ \frac{13}{24} ,[/tex] [tex]n\geq 1[/tex]


Răspuns :

Inductie matematica

a)

pt n=1 verificam: 1/2+1/3+1/4 = 6+4+3  / 12 =13/12>1 ok

pp enuntul adevarat

dem pt n=k+1

1/(k+2) + 1/(k+3) + ... + 1/(3k+4) = 1/(k+1) +1/(k+2) +...+1/(3k+1) - 1/(k+1) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) >(cf ip inductie) 1 - 1/(k+1) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) >

acum ne folosim de inegalitatea 1/(k+1) < 1/(3k+2) si amplificand cu -1 avem

-1/(k+1) > -1/(3k+2) si revenind mai sus, avem:

1 - 1/(k+1) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) > 1 - 1/(3k+2) + 1/(3k+2) + 1/(3k+3) + 1/(3k+4) > 1 + 1/(3k+3) + 1/(3k+4) > 1 deci presupunerea facuta pt n=k este adevarata si pt n=k+1, ∀n,k∈N, n,k≥1.

b)

pt n=1: 1≤1, se verifica

pp adev pt n=k

v.d. pt n=k+1

1+1/3+1/7 +...+1/(2^k -1) + 1/(2^(k+1) - 1) ≤(cf ip ind) k+ p ( si cum p≤1 )⇒≤ k+1

c)

pt n=1: 1/2>13/24≅0,54 FALS, deci enuntul de la c este FALS.