Question

Determinati valorile naturale nenule ale lui n pentru care avem:
0,8(3) < 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/n(n+1) < 0,91(6)
Repede va rog!

Answer 1

1/1*2=1-1/2
1/2*3=1/2-1/3
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1/n(n+1)=1/n-1/(n+1)
deci 1/1*2+1/2*3+.......+1/n(n+1)=1-1/2+1/2-1/3+1/3-1/4+.....+1/n-1/(n+1)=
=1-1/(n+1)=(n+1-1)/(n+1)=n/(n+1)
0,8(3)=75/90=15/18=5/6
0,91(6)=825/900=165/180=33/36=11/12
deci  5/6∠n/(n+1)∠11/12
a) 5/6∠n/(n+1)⇒5n+5∠6n⇒5∠n
b)n/(n+1)/11/12⇒12n∠11n+11⇒n∠11
deci 5∠n∠11⇒n∈(6,7,8,9,10)