.
22
(
)
(
)
∑
∑
−
=
−
−
−
=
−
−
−
=
1
`
0
2
2
0
`
1
'
'
2
'
2
2
m
m
m
m
m
m
m
z
C
z
C
κ
κ
κ
κ
κ
κ
.
Shunday qilib,
(
)
(
)
(
)
∑
−
=
−
−
−
+
+
=
1
0
2
2
2
2
2
2
2
m
m
m
m
m
m
m
m
C
z
z
C
s
co
κ
κ
κ
κ
ϕ
.
Lekin
(
)
(
)
(
)
k
m
s
co
Z
Z
m
m
−
=
+
−
−
−
2
2
2
2
κ
κ
. Shuning uchun
(
)
∑
−
=
+
−
=
1
0
2
2
2
2
2
cos
2
cos
2
m
m
m
m
m
m
C
m
C
κ
κ
ϕ
κ
ϕ
. ■
10-m i s o l. Tenglikni isbotlang:
(
)
(
)
Z
∈
≠
+
=
+
+
+
k
k
in
s
n
in
s
n
in
s
n
in
s
in
s
in
s
,
2
2
2
1
2
...
2
π
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
.
Yechish.
ϕ
ϕ
ϕ
n
s
co
s
co
s
co
A
+
+
+
=
...
2
yig’indini kiritamiz. U holda
isbot qilinayotgan tenglikning chap tomonini B orqali belgilab
(
) (
)
(
)
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
inn
s
i
n
s
co
in
s
i
s
co
in
s
i
s
co
Bi
A
+
+
+
+
+
+
=
+
...
2
2
ni hosil qilamiz.
Bu geometrik progressiya hadlarining yig’indisidan iborat. Quyidagi
belgilashni kiritamiz:
2
sin
2
cos
ϕ
ϕ
α
i
+
=
. U holda
1
...
2
2
2
2
2
4
2
−
−
=
+
+
=
+
+
α
α
α
α
α
α
n
n
Bi
A
.
Oxirgi kasrning surat va maxrajida
α
ning shunday darajalarini qavsdan
tashqariga chiqaramizki, qavs ichida
α
ning qarama-qarshi ko’rsatkichli
darajalarining ayirmasi qolsin (buning mumkin bo’lishi uchun biz
ϕ
ϕ
sin
cos
i
+
ni emas
2
sin
2
cos
ϕ
ϕ
i
+
ni belgilardik):
(
)
(
)
=
−
+
=
+
−
−
+
1
2
α
α
α
α
α
α
n
n
n
Bi
A
(
)
.
2
1
sin
2
1
cos
2
sin
2
sin
2
sin
2
2
sin
2
2
1
sin
2
1
cos
1
1
+
+
+
=
+
+
+
=
−
−
=
−
−
+
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
α
α
α
α
α
n
i
n
n
i
n
i
n
i
n
n
n
n
Bu yerdan
(
)
2
sin
2
1
sin
2
sin
ϕ
ϕ
ϕ
+
=
n
n
B
ni va bir vaqtda
23
(
)
2
2
1
2
ϕ
ϕ
ϕ
in
s
n
cos
n
in
s
A
+
=
ni hosil qilamiz. ■
Xuddi shunday
n
n
b
a
b
a
b
a
cos
...
cos
cos
2
2
1
1
+
+
+
va
n
n
b
a
b
a
b
a
sin
...
sin
sin
2
2
1
1
+
+
+
yig’indilarni ham hisoblash mumkin, agar
n
b
b
b
,...,
,
2
1
argumentlar arifmetik
progressiyani,
n
a
a
a
,...,
,
2
1
koeffisiyentlar esa gometrik progressiyani tashkil
etsa.
M A S H Q L A R
46*. Tengliklarni isbotlang:
a)
2
2
1
2
3
2
1
2
2
...
−
−
=
+
+
+
n
n
n
n
n
C
C
C
, agar n
−
juft bo’lsa;
b)
2
2
1
2
2
2
2
2
...
1
−
−
=
+
+
+
n
n
n
n
C
C
, agar n
−
toq bo’lsa.
47*. Tengliklarni isbotlang:
a)
(
)
−
+
=
⋅⋅
⋅
+
+
+
3
2
2
2
3
1
7
4
1
π
n
s
co
C
C
C
n
n
n
n
;
b)
(
)
−
+
=
⋅⋅
⋅
+
+
+
3
4
2
2
3
1
8
5
2
π
n
s
co
C
C
C
n
n
n
n
.
48*. Quyidagi yig’indini hisoblang:
...
7
5
3
1
+
−
+
−
n
n
n
n
C
C
C
C
.
49*. Ayniyatni isbotlang:
(
)
1
1
1
1
....
4
3
2
1
1
3
4
2
3
1
2
0
+
−
+
=
+
+
+
+
+
+
+
+
n
n
C
C
C
C
C
n
n
n
n
n
n
n
n
κ
κ
κ
κ
κ
κ
.
50*. Ayniyatni isbotlang:
( )
κ
κ
κ
κ
1
2
1
0
1
)
1
...(
−
−
=
−
+
+
−
n
n
n
n
n
C
C
C
C
C
.
51*.
(
)(
)
!
!
-
!
2
...
1
1
0
κ
κ
κ
κ
κ
+
=
+
+
+
−
+
n
n
n
C
C
C
C
C
C
n
n
n
n
n
n
n
n
tenglik
o’rinli
bo’lishini ko’rsating.
52*. Ayniyatni isbotlang:
a)
( ) ( ) ( )
( )
( )
n
n
n
n
n
n
n
n
C
C
C
C
C
2
2
2
2
2
2
2
2
1
2
2
0
2
1
...
−
=
+
+
+
−
;
b)
( ) ( ) ( )
( )
0
...
2
1
2
1
2
2
2
1
2
2
1
1
2
2
0
1
2
=
−
−
+
−
+
+
+
+
+
n
n
n
n
n
C
C
C
C
.
53*. Quyidagi tengliklarni isbotlang:
a)
( )
(
)
∑
−
=
+
+
−
−
=
1
0
2
2
2
2
2
2
1
2
m
m
m
m
m
m
m
C
m
s
co
C
in
s
κ
κ
κ
ϕ
κ
ϕ
;
b)
(
)
∑
=
+
+
+
−
=
m
m
m
m
m
s
co
C
s
co
0
1
2
1
2
2
1
2
2
2
κ
κ
ϕ
κ
ϕ
;
24
c)
( )
(
)
∑
=
+
+
+
+
−
−
=
m
m
m
m
m
m
C
0
1
2
1
2
2
1
2
2
sin
1
sin
2
κ
κ
κ
ϕ
κ
ϕ
.
54*. Tengliklarni isbotlang:
a)
(
)
0
1
2
cos
...
5
cos
3
cos
cos
=
−
+
+
+
+
n
n
n
n
n
π
π
π
π
;
b)
(
)
0
1
2
sin
...
5
sin
3
sin
sin
=
−
+
+
+
+
n
n
n
n
n
π
π
π
π
.
55*. Yig’indilarni toping:
a)
(
)
x
n
C
x
C
x
n
n
n
1
cos
...
2
cos
cos
1
+
+
+
+
;
b)
(
)
x
n
C
x
C
x
n
n
n
1
sin
...
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