Question

$\frac{1}{n+1} + \frac{1}{n+2} +.... + \frac{1}{3n+1} \ \textgreater \ 1$
Trebuie sa o demonstrez prin inductia matematica

Answer 1

Notam P(n) inegalitatea 1/(n+1)+1(n+2)+...+1/(3n+1) > 1

1) P(n) adevarata?

Pentru n=1 => P(2): 1/(2+1)+1(2+2)>1 => 1/3+1/4>1 => 7/12>1/12 (A)

2) P(k) => P(k+1)

P(n+1)=

1/(n+2)+1/(n+3)+...+1/(3n+1)+1/(3n+2)+1/(3n+3)+1/(3n+4) > 1-1/(n+1)+1/(3n+2)+1/(3n+3)+1/(3n+4)

= 1+ ceva/[3(n+1)(3n+2)(3n+4)] >1

ceva>0 ; 3(n+1)(3n+2)(3n+4)>0 => (A)

Din (1) si (2) => P(n) adevarata oricare ar fi n>1