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1
2n+ 1
\u2212
1
2n+ 3
. . . c.q.d.
12) Calcular a soma (S) gerada por
1
3
+
1
3× 5 +
1
5× 7 + · · ·+
1
(2n+ 1)(2n+ 3)
Resoluc¸a\u2dco:
S =
1
3
+
1
3× 5 +
1
5× 7 + · · ·+
1
(2n+ 1)(2n+ 3)
2S =
2
3
+
2
3× 5 +
2
5× 7 + · · ·+
2
(2n+ 1)(2n+ 3)
Sabemos que
2
(2n+ 1)(2n+ 3)
=
1
2n+ 1
\u2212
1
2n+ 3
, onde n pode ser:
0, 1, 2, . . ., k.
Se n = 0\u2192 2
3
=
1
1
\u2212
1
3
Se n = 1\u2192 2
3× 5 =
1
3
\u2212
1
5
Se n = 2\u2192 2
5× 7 =
1
5
\u2212
1
7
...
...
Se n = k\u2192 2
(2k+ 1)(2k+ 2)
=
1
2k+ 1
\u2212
1
2k+ 3
\u201cMain\u201d
2006/12/15
page 284
284 [CAP. 7: NU´MEROS FRACIONA´RIOS
Somando membro a membro, teremos:
2S =
1
1
\u2212
1
2k+ 3
2S =
2k+ 3\u2212 1
2k+ 3
S =
2(k+ 1)
2(2k+ 3)
S =
k+ 1
2k+ 3
Como k = n, teremos :
S =
n+ 1
2n+ 3
13) Se A = 111 . . .11\ufe38 \ufe37\ufe37 \ufe38
8 algs
e B = 111 . . .11\ufe38 \ufe37\ufe37 \ufe38
100 algs
, calcular o mdc de A e B.
Resoluc¸a\u2dco:
A =
108 \u2212 1
9
e B =
10100 \u2212 1
9
m.d.c.
(
108 \u2212 1
9
,
10100 \u2212 1
9
)
= m
m.d.c
(
9× 10
8 \u2212 1
9
, 9× 10
100 \u2212 1
9
)
= 9×m.
m.d.c. (108 \u2212 1, 10100 \u2212 1) = 9×m.
Sabemos da A´lgebra que: am \u2212 bm e´ divis´\u131vel por (a \u2212 b).
Observe que (108 \u2212 1) na\u2dco divide 10100 \u2212 1.
Fatorando, enta\u2dco, 108 \u2212 1, teremos:
10100 \u2212 1
108 \u2212 1
=
(104)25 \u2212 (1)25
(104 + 1)(104 \u2212 1)
.
Ve\u2c6-se pois que (104 \u2212 1) divide (104)25 \u2212 (1)25, da´\u131 ...
9×M = 104 \u2212 1
M =
104 \u2212 1
9
= 1.111.
\u201cMain\u201d
2006/12/15
page 285
[SEC. 7.21: EXERC´ICIOS RESOLVIDOS 285
Logo, o m.d.c.
(
108 \u2212 1
9
,
10100 \u2212 1
9
)
= 1.111.
14) Tre\u2c6s torneiras enchem um tanque do seguinte modo: a 1a mais a 3a em
84 minutos; a 1a mais a 2a em 70 minutos, e a 2a mais a 3a em 140
minutos. Achar o tempo que levara´ cada uma para enche\u2c6-lo.
Resoluc¸a\u2dco:
Chamemos as 1a , 2a e 3a torneiras, de A, B e C, respectivamente.
Sendo A +C = 84 min., A+ B = 70 min. e B+ C = 140 min.
Logo, a cada minuto, a parte do tanque enchido e´:
A+ C =
1
84
. . . (I)
A+ B =
1
70
. . . (II)
B +C =
1
140
. . . (III)
Somando-se I, II e III, teremos: 2× (A + B +C) = 1
84
+
1
70
+
1
140
· .
Simplificando, vira´:
A+ B+ C =
1
60
Substituindo (III) em (IV) . . .A =
1
60
\u2212
1
140
=
4
420
=
1
105
Substituindo (I) em (IV) . . .B =
1
60
\u2212
1
84
=
2
240
=
1
210
Substituindo (II) em (IV) . . .C =
1
60
\u2212
1
70
=
1
420
tanque min.
1a Torneira:
1
105
. . . 1
1 . . . 105
tanque min.
2a Torneira:
1
210
. . . 1
1 . . . 210
\u201cMain\u201d
2006/12/15
page 286
286 [CAP. 7: NU´MEROS FRACIONA´RIOS
tanque min.
3a Torneira:
1
420
. . . 1
1 . . . 420
Resp.:
1a : 105 min.
2a : 210 min.
3a : 420 min.
15) Um cachorro persegue uma lebre; enquanto o cachorro da´ 4 pulos, a
lebre da´ 9, pore´m, 2 pulos do cachorro valem 7 pulos da lebre. Sendo
a dista\u2c6ncia entre os dois de 100 pulos de lebre, determinar o nu´mero de
pulos o cachorro devera´ dar para alcanc¸ar a lebre.
1a resoluc¸a\u2dco:
Cachorro Lebre
4 ........ 9
2 ........ 7
4 ........ 9
4 ........ 14
Ve\u2c6-se que a cada 4 pulos o cachorro avanc¸a 5(14\u2212 9) pulinhos de lebre.
Cach avanc¸o
4 \u2212\u2192 5
80 \u2190\u2212 100
Resp.:80 pulos
2a resoluc¸a\u2dco: (alge´brica)
Seja p o nu´mero de pulos que o cachorro deve dar para alcanc¸ar a lebre.
Do que foi desenvolvido anteriormente, podemos simplesmente escrever
a equac¸a\u2dco:
7
2
p\u2212
9
4
p = 100 \u2234 p = 80
\u201cMain\u201d
2006/12/15
page 287
[SEC. 7.22: EXERC´ICIOS PROPOSTOS 287
7.22 Exerc´\u131cios Propostos
1) Efetue e simplifique as frac¸o\u2dces, ate´ torna´-las irredut´\u131veis:
a)
2
7
+
3
7
b)
7
9
\u2212
1
9
\u2212
2
9
c)
1
4
+
1
2
d)
1
3
\u2212
1
9
e)
1
7
+
1
3
f)
1
4
\u2212
1
5
g)1+
3
2
h) 1\u2212
1
5
i) 2× 3
5
j)
2
3
+
3
4
k) 2+
3
5
l)
4
5
+ 3
m)
2
2
5
n)
1+
1
2
1\u2212
1
2
o)
1
3
+
1
12
+
5
8
1
6
+
1
3
+ 4+
1
2
p)
5\u2212
1
10(
1+
1
2
)
+
(
3+
2
5
)
q)
1
12
+
1
9
× 3
4
7
12
\u2212
1
4
+
3
4
+
3
8
+
3
16
+
9
4
11
16
\u2212
1
4
× 9
4
r) 1\u2212
4\u2212
1
3
× (1+ 1
5
)
1
8
+
139
40
s)
3
8
+
1
4
1+
1+
1
4
4
t)
1
2+
3
4+
5
6
\u201cMain\u201d
2006/12/15
page 288
288 [CAP. 7: NU´MEROS FRACIONA´RIOS
u)
1
4+
1
2+
1
3+
1
5
v) 1+
1
1\u2212
1
1+
1
1\u2212
1
2
w)
1
1+
1
1+
1
1+
1
2
x)
1\u2212
1
3
1+
1
3
+ 1+
1+
1
2
1\u2212
1
2
4+
1
2
y) 1\u2212
1\u2212
1\u2212
1
2
1+
1
2
1+
1+
1
2
1\u2212
1
2
z)
3
1+
1
1\u2212
1
2
1+
1
1+
1
2
3
+
1
2
5
1\u2212
1
5
\u2212 1
2) Efetue e simplifique as frac¸o\u2dces, ate´ torna´-las irredut´\u131veis:
a)
(
1+
1
2
)
×
(
1+
1
3
)
×
(
1+
1
4
)
× · · · ×
(
1+
1
10
)
;
b)
(
1\u2212
2
3
)
×
(
1\u2212
2
4
)
×
(
1\u2212
2
5
)
× · · · ×
(
1\u2212
2
99
)
×
(
1\u2212
2
100
)
;
c)
(
1+
1
15
)
×
(
1+
1
16
)
×
(
1+
1
17
)
×
(
1+
1
18
)
×
(
1+
1
19
)
;
d)
(
1\u2212
3
7
)
×
(
1\u2212
3
8
)
×
(
1\u2212
3
9
)
× · · · ×
(
1\u2212
3
20
)
;
e)
(
1\u2212
1
2
)
×
(
1\u2212
1
3
)
×
(
1\u2212
1
4
)
× · · · ×
(
1\u2212
1
99
)
×
(
1\u2212
1
100
)
;
f)
22 \u2212 1
22 + 2
× 3
2 \u2212 1
32 + 3
× 4
2 \u2212 1
42 + 4
× · · · × 19
2 \u2212 1
192 + 19
× 20
2 \u2212 1
202 + 20
;
\u201cMain\u201d
2006/12/15
page 289
[SEC. 7.22: EXERC´ICIOS PROPOSTOS 289
g) 2×
(
1\u2212
1
2
)
+ 3×
(
1\u2212
1
3
)
+ 4×
(
1\u2212
1
4
)
+ · · ·+ 10×
(
1\u2212
1
10
)
;
h) 20+
(
20+
1
5
)
+
(
20+
2
5
)
+
(
20+
3
5
)
+ · · ·+ 40;
i)
1
2
+
(
1
3
+
2
3
)
+
(
1
4
+
2
4
+
3
4
)
+ · · ·+
(
1
100
+
2
100
+ · · ·+ 98
100
+
99
100
)
;
j)
1
2
\u2212
1
3
1
3
\u2212
1
4
×
1
4
\u2212
1
5
1
5
\u2212
1
6
×
1
6
\u2212
1
7
1
7
\u2212
1
8
× · · · ×
1
2004
\u2212
1
2005
1
2005
\u2212
1
2006
;
k)
1
1× 3 +
1
3× 5 +
1
5× 7 +
1
7× 9 +
1
9× 11 + · · ·+
1
97× 99 ;
l)
2
1× 3 +
2
3× 5 +
2
5× 7 + · · ·+
2
19× 21 ;
m)
1
22 \u2212 1
+
1
32 \u2212 1
+
1
42 \u2212 1
+ · · ·+ 1
92 \u2212 1
+
1
102 \u2212 1
;
n)
(
1\u2212
1
22
)
×
(
1\u2212
1
32
)
×
(
1\u2212
1
42
)
×· · ·×
(
1\u2212
1
992
)
×
(
1\u2212
1
1002
)
;
o)
1
1× 4 +
1
4× 7 +
1
7× 10 + · · ·+
1
28× 31 ;
p)
13
2× 4 +
13
4× 6 +
13
6× 8 + · · ·+
13
50× 52 ;
q)
(
14 +
1
4
)
×
(
34 +
1
4
)
×
(
54 +
1
4
)
× · · · ×
(
114 +
1
4
)
(
24 +
1
4
)
×
(
44 +
1
4
)
×
(
64 +
1
4
)
× · · · ×
(
124 +
1
4
) ;
r)
1
10
+
1
18
+
1
28
+ · · · (Harvard)
3) Calcule:
a) 16
1
2
b) 8
1
3
\u201cMain\u201d
2006/12/15
page 290
290 [CAP. 7: NU´MEROS FRACIONA´RIOS
c) 49
1
2
d) 64
1
3
e) 81
1
4
f) 625
1
4
g)
(
810 + 410
84 + 411
)1
2
4) Se
43
19
= a +
1
b +
1
c+
1
d
, calcule a + b + c + d.
5) Determine a+ b+ c, sabendo que a +
1
b +
1
c
=
15
2
·
6) Se A, B e C sa\u2dco inteiros positivos, onde
31
4
= A +
1
B+
1
C+ 1
, calcule
A2 + B2 + C2.
7) Considere os nu´meros a =
105678+ 10999
105679+ 10999
e b =
105679+ 10999