|
“O
Bilim
Ta’li
Ta’li
OLIY M
IQTI
m sohasi:
im sohasi:
im yo‘nalis
OLIY V
T
ATEMA
ISODC
fani
10
20
23
shlari: 52
52
52
52
52
52
52
52
52
52
O‘ZBEK
VA O‘RTA
TOSHKE
ATIKA,
CHILA
idan o’
Amal
00000
–
00000
–
0000
–
30600
–
30700
–
30800
–
30900
–
31200
–
31300
–
31500
–
32000
–
32700
–
32800
–
KISTON R
A MAXSU
ENT MOL
STATIS
KAFED
AR UCH
’quv-u
(II sem
liy mas
Toshkent
– Gumanita
– Ijtimoiy s
– Iqtisod
– Moliya;
– Bank ishi
– Soliqlar v
– Buxgalter
– Sug‘urta i
– Pensiya is
– Baholash
– Davlat bu
– Investitsio
– Elektron t
RESPUBL
US TA’LI
LIYA INS
STIKA V
DRASI
HUN M
uslubiy
mestr)
shg‘ulo
t – 2020
ar
oha, iqtisod
;
va soliqqa to
riya hisobi v
ishi;
shi;
ishi;
udjetining g‘
on loyihalar
tijorat.
LIKASI
IM VAZI
STITUTI
VA EKO
MATE
majm
otlar
d va huquq
ortish;
va audit (ta
‘azna ijrosi;
rni moliyala
IRLIGI
ONOME
EMATI
ua
armoqlar bo
;
ashtirish;
TRIKA”
IKA
‘yicha);
”
19-amaliy mashg‘ulot.
n
fazoda nuqtalarning o‘zaro joylashishi.
Sonli ketma-ketlik
Quyidagisohalarbilanchegaralanganto‘plamlarqavariqto‘plambo‘ladimi?
19.1.
=
+
=
+
17
6
6
8
3
3
2
1
2
1
x
x
x
x
19.2.
≤
+
−
≤
+
2
2
2
1
2
1
x
x
x
x
19.3.
≥
≥
≤
+
≤
+
0
,
0
30
6
5
20
5
2
2
1
2
1
2
1
x
x
x
x
x
x
19.4.
≥
≥
≥
−
≤
+
−
0
,
0
1
2
1
2
2
1
2
1
2
1
x
x
x
x
x
x
19.5.
≤
+
−
≤
+
0
1
3
1
6
2
2
1
2
2
2
1
x
x
x
x
19.6.
5
3
2
2
2
1
≤
+ x
x
Quyidagito‘plamlarqavariqmi?
19.7. Markazsizdoira. Markazsizshar.
19.8.Kesmavabukesmadayotmaydigannuqta.
19.9.R
3
- fazodaumumiynuqtagaegabo‘lgan 2 ta tetraedr.
19.10.R
2
- fazodaumumiytomongaegabo‘lgan 2 ta uchburchak.
19.11.
19.12.
19.13.19.14.
19.15.
19.16.
Sonliketma-ketlikchegaralanganliginiisbotlang:
Isbot:
va
shuninguchun
.
Sonliketma-ketlikchegaralanganliginiisbotlang.
19.17.
19.18
19.19.
19.20.
Sonliketma-ketlikmonotonliginiisbotlang:
Isbot:
.
Demak,
shuninguchunbuketma-
ketlikmonotonkamayuvchi.
Sonliketma-ketlikmonotonliginiisbotlang:
19.21.
19.23.
19.24.
19.25.
Ketma-ketliklimitita’rifidanfoydalanib, quyidagilarniisbotlang:
Isbot: ixtiyoriy
son olamiz,
tengsizlikniqanoatlantiruvchi
larnitopishuchun
tengsizlikniyechamiz.
2
1
2
2
+
+
=
n
n
x
n
2
1
1
2
1
2
2
2
+
−
=
+
+
n
n
n
2
1
2
1
0
2
≤
+
<
n
1
2
1
<
<
n
x
( )
2
1
1
2
+
+
−
=
n
n
x
n
n
n
Sin
x
n
=
)
)
1
(
1
(
n
n
x
−
−
=
)
1
(
),
1
lg(
lg
>
−
−
=
n
n
n
x
n
1
lg
)
1
lg(
lg
−
=
−
−
=
n
n
n
n
x
n
0
)
1
1
lg(
1
lg
1
lg
1
lg
2
2
2
1
<
−
=
−
=
−
−
+
=
−
+
n
n
n
n
n
n
n
x
x
n
n
,
,
0
1
1
n
n
n
n
x
x
x
x
<
<
−
+
+
n
n
n
x
2
3
−
=
1
2
−
= n
x
n
∑
=
=
n
k
n
k
x
1
1
1
lim
=
+
∞
→
n
n
n
0
>
ε
ε
<
−
+
=
−
+
=
−
1
;
1
1
1
1
1
n
n
x
n
n
n
x
n
1
1
n
ε
<
+
. Shundayqilib,
soniningbutunqismi
boʻladi, u holda
tengsizlikbarcha
lardabajariladi.
-ixtiyoriysonbo
ʻlganiuchun
Agar
bo
ʻlsa,
larda
bo
ʻladi.
Ketma-ketliklimitita’rifidanfoydalanib, quyidagilarniisbotlang:
19.26.
19.26.
19.27.
qaysi
danboshlab,
tengsizlikoʻrinliboʻladi?
Umumiyhadiorqaliberilganketma-ketlikningbirinchibeshtahadiniyozing:
19.28.
19.29.
19.30.
Ketma-
ketlikningberilganhadlariorqaliumumiyhadiningformulasiniyozing:
19.31.
19.32.
19.33. 2; 10; 26; 82; 242; 730; …
Quyidagilimitlarnitoping:
19.34.
19.35.
19.36.
19.37.
19.38.
19.39.
ε
ε
−
>
1
n
ε
ε
−
1
−
=
ε
ε
1
N
ε
<
−1
n
x
N
n
>
ε
.
1
1
lim
=
+
∞
→
n
n
n
01
.
0
=
ε
99
,
99
01
.
0
01
.
0
1
>
=
−
=
n
N
01
.
0
1
<
−
n
x
2
1
2
1
4
lim
=
+
−
∞
→
n
n
n
5
3
1
5
1
3
lim
=
−
+
∞
→
n
n
n
3
2
3
2
1
2
lim
−
=
−
−
∞
→
n
n
n
n
0001
.
0
3
2
3
2
1
2
<
−
−
−
−
n
n
1
2
1
+
=
n
x
n
1
2
3
+
+
=
n
n
x
n
2
1
)
1
(
n
n
x
n
n
+
−
=
...
;
24
1
;
6
1
;
2
1
;
1
...
;
25
6
3
;
16
1
3
;
9
7
2
;
4
1
2
;
1
1
4
2
3
3
3
lim
−
+
∞
→
n
n
n
1
3
2
3
3
lim
−
+
+
∞
→
n
n
n
n
1
5
)
1
(
3
3
lim
+
+
∞
→
n
n
n
n
n
n
n
+
+
∞
→
4
3
2
lim
1
2
1
2
2
2
4
lim
−
+
+
+
∞
→
n
n
n
n
n
!
)!
1
(
!
lim
n
n
n
n
−
+
∞
→
19.40.
19.41.
19.42.
19.43.
19.44.
19.45.
19.46.
19.47.
n
n
n
3
1
...
3
1
1
2
1
...
2
1
1
lim
+
+
+
+
+
+
∞
→
4
3
...
9
6
3
2
lim
+
+
+
+
+
∞
→
n
n
n
)
3
2
(
lim
−
−
+
∞
→
n
n
n
n
)!
2
2
(
)!
3
2
(
)!
2
2
(
)!
1
2
(
lim
+
−
+
+
+
+
∞
→
n
n
n
n
n
(
)
3
3
3
8
2
1
lim
n
n
n
n
→∞
+
+ −
−
−
+
+
+
∞
→
1
4
1
...
15
1
3
1
2
lim
n
n
3
1
3
2
6
1
lim
n
n
Cos n
n
n
→∞
−
+
−
+
+
∞
→
2
2
2
9
1
2
!
1
lim
n
n
n
Sin
n
n
n
20-amaliy mashg‘ulot.Sonli qatorlar
20.1. Ushbu qatorni yaqinlashishga tekshiring:
1
1
1
1
1 3
2 4
3 5
(
2)
n n
+
+
+
+
+
⋅
⋅
⋅
+
.
Yechish. Berilgan qatorning n-xususiy yig‘indisi
1
1
1
1
1 3
2 4
3 5
(
2)
n
S
n n
=
+
+
+
+
⋅
⋅
⋅
+
. Bu yig‘indini soddalashtirish
maqsadida qatorning n-hadini quyidagi
1
1 1
1
(
2)
2
2
n n
n
n
=
−
+
+
ko‘rinishda yozib
olamiz. U holda
1 1
1
1 1
1
1 1
1
1
1
1
1 1
1
2 1
3
2 2
4
2 3
5
2
1
1
2
2
n
S
n
n
n
n
=
−
+
−
+
−
+
+
−
+
−
−
+
+
=
=
1
1
1
1
1
2
2
1
2
n
n
+ −
−
+
+
bo‘ladi. Ravshanki, { S
n
} ketma-ketlik limiti mavjud va
3
4
ga teng. Demak, berilgan qator yaqinlashuvchi bo‘lib, uni
3
4
=
1
1
1
1
1 3
2 4
3 5
(
2)
n n
+
+
+
+
+
⋅
⋅
⋅
+
, yoki
3
4
=
1
1
(
2)
n
n n
∞
=
+
∑
kabi yozish mumkin
ekan.
20.2. Ushbu qatorning n-xususiy yig‘indisi uchun ifoda toping va yaqinlashishga
tekshiring:
a)
1
2
(2
1)(2
1)
n
n
n
∞
=
−
+
∑
; b)
1
2
3
5
n
n
n
n
∞
=
+
∑
.
20.3.
sonli qatorning yaqinlashuvchi
bo‘lishining zaruriy sharti bajarilishini tekshiring.
Yechish.
= ;
=
. Demak yaqinlashuvchi
bo‘lishining zaruriy sharti bajarilmaydi, qator uzoqlashuvchi.
...
3
5
1
2
...
12
7
7
5
2
3
3
5
1
2
1
+
−
+
+
+
+
+
=
−
+
∑
∞
=
n
n
n
n
n
3
5
1
2
−
+
=
n
n
a
n
lim
∞
→
n
3
5
3
2
−
+
n
n
5
2
n
n
a
lim
∞
→
0
5
2 ≠
20.4.Qatorlar uchun yaqinlashuvchi bo‘lishining zaruriy sharti bajarilishini
tekshiring:
a)
1
1
.
2
1
n
n
n
∞
=
+
+
∑
b)
(
)
1
2
.
ln
1
n
n
n
∞
=
+
+
∑
20.5. Birinchi taqqoslash alomatidan foydalanib
...
n
...
n
+
+
+
+
+
3
2
1
3
2
3
1
3
2
2
1
3
2
3
2
qatorni yaqinlashishga tekshiring.
Yechish. Ushbu qatorni qaraymiz:
...
...
n
+
+
+
+
+
3
2
3
2
3
2
3
2
3
2
.
Ravshanki,
n
n
n
n
b
n
a
=
≤
=
3
2
3
2
1
. Mahraji
2
3
q
=
bo‘lgan
∑
∞
=
1
3
2
n
n
geometrik qator yaqinlashuvchi, demak yuqoridagi teoremaga ko‘ra berilgan
1
1
2
3
n
n
n
∞
=
⋅
∑
qator ham yaqinlashuvchi bo‘ladi.
Taqqoslash alomatini qo‘llab berilgan qatorlarni yaqinlashishga tekshiring:
20.6.
2
1
2
.
3
5
n
n
n
∞
=
−
∑
20.7.
3
1
2
7
.
3
11
n
n
n
∞
=
+
+
∑
20.8.
2
5
1
7
.
12
n
n
n
∞
=
+
+
∑
20.9.
(
)
1
1
.
ln
3
n
n
∞
=
+
∑
20.10.
(
)
1
3
.
3
4
n
n
n
n
∞
=
−
∑
20.11.
(
)
2
1
ln
3
.
n
n
n
∞
=
+
∑
20.12. Qatorni yaqinlashishga tekshiring:
...
n
...
n
+
+
+
+
+
2
2
3
2
2
2
2
3
2
2
2
1
2
Yechish.Ravshanki,
2
2
n
a
n
n
=
,
2
1
1
1
2
)
n
(
a
n
n
+
=
+
+
.
1
2
2
1
2
2
2
(
1)
lim
lim
2lim
2 1
2
(
1)
n
n
n
n
n
n
n
a
n
n
a
n
n
+
+
→∞
→∞
→∞
+
=
=
= >
+
.
Demak, qatoruzoqlashuvchi.
Dalamber alomati yordamida qatorni yaqinlashishga tekshirish mumkinmi?
20.13.
∑
∞
=
+
1
7
3
2
n
n
n
20.14.
∑
∞
=
+
⋅
1
12
5
2
n
n
n
n
20.15.
(
)
∑
∞
=
+
1
1
3
!
n
n
n
n
20.16.
∑
∞
=
−
+
1
3
2
5
1
n
n
n
20.17.
∑
∞
=
+
1
2
2
n
n
n
n
20.18.
∑
∞
=
⋅
1
3
!
n
n
n
n
n
20.19.
∑
∞
=
+
1
2
5
!
n
n
n
n
20.20.
(
)
∑
∞
=
+
+
1
2
!
1
n
n
n
n
n
20.21. Berilgan qatorni yaqinlashishga tekshiring:
...
)
n
(
ln
...
ln
ln
n
+
+
+
+
+
1
1
3
1
2
1
2
Yechish.
1
1
lim
lim
lim
0 1
ln (
1)
ln(
1)
n
n
n
n
n
n
n
a
n
n
→∞
→∞
→∞
=
=
= <
+
+
Demak, qatoryaqinlashuvchi.
Koshiningradikal
alomati yordamida qatorlarning yaqinlashishishni
tekshiring:
20.22.
2
1
1
1
.
n
n
n
∞
=
+
∑
20.23.
1
1
.
2
1
n
n
n
n
∞
=
−
+
∑
20.24.
2
1
2
1
.
3
1
n
n
n
n
∞
=
−
+
∑
20.25.
(
1)
1
1
.
1
n n
n
n
n
−
∞
=
−
+
∑
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