Question

Demonstrati prin inductie matematica ca:
$1 \div (n + 1) + 1 \div ( n+ 2) + ... + 1 \div 2n > 13 \div 24$
oricare ar fi n
$n\geqslant 1$

Answer 1

pt n=1
1/2=12/24>13/24 fals
sau usor interpretabil pt ca n+1=2n, dac n=1 si atunci ; luam un termen, sau 2??
dac luam 2 termeni
1/2+1/2=1>13/24
pt rigoare prefer insa sa luam de la
 n=2
1/3+1/4=4/12+3/12=7/12=14/24 adevarat
presupunem adevarat pt n, n≥2 si notam suma cu Sn
pt n->n+1

atunci S(n+1)=1/(n+2)+1/(n+3)+...1/(2n+2)=

=-1/(n+1)+(1/(n+1)+1/(n+2)+...+1/2n)+1/(2n+1)+1/(2n+2)=

=Sn+1/(2n+1)+1/(2n+2)-1/(n+1)=Sn+1/(2n+1)-1/(2n+2)
dar Sn>13/24 presupusa adevarata si verificat pt n=2
si 1/(2n+1)-1(2n+2)=Δ>0 pt ca 0<2n+1<2n+2

deci Sn+Δ>Sn>13/24
Pn->P(n+1) am verificat ficat! ficat! prin inductie matematica (paranteza completa)