[*رتاج الكعبة*]

مادة الرياضيات

ﺍﻟﺘﻤﺭﻴﻥ ﺍﻷﻭل: (05 ﻨﻘﺎﻁ)
ﻋﻠﻰ1 n .5) ﻋﻴﻥ ﺤﺴﺏ ﻗﻴﻡ ﺍﻟﻌﺩﺩ ﺍﻟﻁﺒﻴﻌﻲ n ﺒﻭﺍﻗﻲ ﺍﻟﻘﺴﻤﺔ ﺍﻹﻗﻠﻴﺩﻴﺔ ﻟﻠﻌﺩﺩ 3
2013 ﻋﻠﻰ2 2009 .5) ﻋﻴﻥ ﺒﺎﻗﻲ ﻗﺴﻤﺔ ﺍﻟﻌﺩﺩ 2013 ﻋﻠﻰ 5 ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﺒﺎﻗﻲ ﻗﺴﻤﺔ ﺍﻟﻌﺩﺩ
+ 4 128 ×2013 ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ3 2009 .5) ﺍﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻌﺩﺩ

ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻨﻲ: (05 ﻨﻘﺎﻁ)
≡ . 1 n n) ﺃ) ﺒﺭﻫﻥ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ 9 2 [7] ، n
ﺃﻥ ﺒﺭﺭ (ﺏ n n 3 9 2 =
ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ 2 2 1 2 3 2 + + + n n . 7) ﺍﺴﺘﻨﺘﺞ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ n :
693 ﻋﻠﻰ 3 2013 . 7) ﻋﻴﻥ ﺒﺎﻗﻲ ﺍﻟﻘﺴﻤﺔ ﺍﻹﻗﻠﻴﺩﻴﺔ ﻟﻠﻌﺩﺩ

( ) (ﻨﻘﺎﻁ 04) :ﺍﻟﺜﺎﻟﺙ ﺍﻟﺘﻤﺭﻴﻥ n : ﻭﺘﺤﻘﻕ r ﺃﺴﺎﺴﻬﺎ ﺤﺴﺎﺒﻴﺔ ﻤﺘﺘﺎﻟﻴﺔ u 1 ﻭ u =13 3 5 . u u+ =176
1 0 . u) ﺍﺤﺴﺏ ﺍﻷﺴﺎﺱ r ﻭ ﺍﻟﺤﺩ ﺍﻷﻭل
u ﺒﺩﻻﻟﺔ 2 n . n) ﺃ) ﻋﻴﻥ ﻋﺒﺎﺭﺓ
u ﺍﻟﻤﺴﺎﻭﻱ ﻟـ n . 2013 ﺏ) ﻤﺎ ﻫﻭ ﺍﻟﺤﺩ ﻤﻥ ( )
ﻨﻀﻊ (3 0 1 .... n n . Sn =1243 ﻴﻜﻭﻥ ﺤﺘﻰ n ﺍﻟﻁﺒﻴﻌﻲ ﺍﻟﻌﺩﺩ ﻋﻴﻥ . S uu u = ++ +

ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺭﺍﺒﻊ: (06 ﻨﻘﺎﻁ)
u ﻤﺘﺘﺎﻟﻴﺔ ﻋﺩﺩﻴﺔ ﻤﻌﺭﻓﺔ ﺒﺤﺩﻫﺎ ﺍﻷﻭل u0 = 2 ﻭﻤﻥ ﺍﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ n un+1 = 2un −1 ، n ﻟﺘﻜﻥ ( )
ﺍﺤﺴﺏ (1 1 ، u 2 ﻭ u u3 .
v ﻜﻤﺎ ﻴﻠﻲ : 2 n .vn = un −1) ﻨﻌﺭﻑ ﻤﻥ ﺍﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ nﺍﻟﻤﺘﺘﺎﻟﻴﺔ ( )
u ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻴﻁﻠﺏ ﺘﻌﻴﻴﻥ ﺃﺴﺎﺴﻬﺎ ﻭﺤﺩﻫﺎ ﺍﻷﻭل . n ﺃ) ﺍﺜﺒﺕ ﺃﻥ ( )
ﺍﻜﺘﺏ (ﺏ n ﺜﻡ n ﺒﺩﻻﻟﺔ v n .n ﺒﺩﻻﻟﺔ u
u ﺜﻡ ﺤﺩﺩ ﺭﺘﺒﺘﻪ . n ﺝ) ﺍﺜﺒﺕ ﺃﻥ ﺍﻟﻌﺩﺩ 65 ﺤﺩ ﻤﻥ ﺤﺩﻭﺩ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ( )
ﺍﻟﻤﺠﻤﻭﻉ ﺍﺤﺴﺏ (ﺩ 0 1 2 6 .S

الحل

ﺍﻟﺘﻤﺭﻴﻥ ﺍﻷﻭل:
1 n : 5) ﺒﻭﺍﻗﻲ ﺍﻟﻘﺴﻤﺔ ﺍﻹﻗﻠﻴﺩﻴﺔ ﻟﻠﻌﺩﺩ 3
[ ] 4 3 15 k [ ] ، ≡ 4 1 3 35 k + [ ] ، ≡ 4 2 3 45 k + [ ] ، ≡ 4 3 3 25 k + . ≡
. 3 ﻫﻭ 5 ﻋﻠﻰ 2013 ﺍﻟﻌﺩﺩ ﻗﺴﻤﺔ ﺒﺎﻗﻲ ﺇﺫﻥ2013 5 402 3 = × + (2
[ ] ﺃﻥ ﻨﺴﺘﻨﺘﺞ 2009 4 502 1 = × + ﻭ 2013≡3 5[ ] 2009 2013 ≡3 5
2013 ﻋﻠﻰ5 ﻫﻭ 2009 . 3 ﺇ ﺫﻥ ﺒﺎﻗﻲ ﻗﺴﻤﺔ ﺍﻟﻌﺩﺩ
(3 [ ]
[ ]
2009 4 128 4 3 3 5 2013
0 5
× + ≡×+
≡
+ 4 128 ×2013 ﻴﻘﺒل ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ2009 .5 ﻨﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻌﺩﺩ


ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻨﻲ



9 2 [ ] 7 ﻭﻤﻨﻪ 9 27 ≡ [ ] ﻟﺩﻴﻨﺎ n ﻁﺒﻴﻌﻲ ﻋﺩﺩ ﻜل ﺃﺠل ﻤﻥ (ﺃ (1 n n . ≡
ﻟﺩﻴﻨﺎ (ﺏ 2 ( ) ﻭﻤﻨﻪ 3 9 = 2 3 9 n n ﺃﻱ = n n 3 9 2 . =
(2
[ ]
[ ]
[ ]
21 2 2 3 2 3 3 2 4 9 3 9 47
9 77
0 7
nn n n n n
n
+ + + = ×+ × ≡ ×+ ×
≡ ×
≡

ﻭﻤﻨﻪ 2 1 2 3 2 + + + n n . 7 ﻋﻠﻰ ﺍﻟﻘﺴﻤﺔ ﻴﻘﺒل
[ ]ﺃﻱ 693 0 7 ≡ [ ] ﻭﻤﻨﻪ 693 7 999 = × (3 2013 693 0 7 ≡
693 ﻋﻠﻰ 7 ﻫﻭ 2013 . 0 ﺇﺫﻥ ﺒﺎﻗﻲ ﺍﻟﻘﺴﻤﺔ ﺍﻹﻗﻠﻴﺩﻴﺔ ﻟﻠﻌﺩﺩ


ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻟﺙ:


. u u 3 5 + =176ﻭ u1 =13 : ﻭﺘﺤﻘﻕ r ﺃﺴﺎﺴﻬﺎ ﺤﺴﺎﺒﻴﺔ ﻤﺘﺘﺎﻟﻴﺔ(un )
(1 3 5 ﺘﻜﺎﻓﺊ u u+ =176 1 1 u ru r + ++ = 2 4 176
r = 25 ﻭﻴﻜﻭﻥ 6 150 r = ﺃﻱ 26 6 176 + r = ﻭﻤﻨﻪ
1 0 u0 = −12 ﺃﻱ 13 25 =u0 + ﻭﻤﻨﻪ uu r = +
ﻋﺒﺎﺭﺓ (ﺃ (2 n : n ﺒﺩﻻﻟﺔ u n 0 u n n =− + 12 25 ﻭﻤﻨﻪ u u nr = +
2013 (ﺏ n 25 2025 n = ﻭﻤﻨﻪ −12 25 2013 + = n ﺘﻜﺎﻓﺊ u =

u ﺍﻟﻤﺴﺎﻭﻱ ﻟـ 2013 ﻫﻭ n ﺃﻱn =81 . ﺇﺫﻥ ﺍﻟﺤﺩ ﻤﻥ ( )
01 0 ( ) ( ) (3 1 1 .... 25 24
2 2 n nn
n n S uu u uu n + + . =++ + = + = −
1243 n ( ) ﺘﻜﺎﻓﺊ S = 1 25 24 1243
2
n
n
+ − =
ﻭﻤﻨﻪ 2 25 22510 0 n n + − =
( )
2
،∆= = 251001 501 1 n2 = ∈ 10 ` ﻭ n = − ∉ 10,04 `
n =10 ﺇﺫﻥ

2 un+1 = 2un −1 ، n ﻁﺒﻴﻌﻲ ﻋﺩﺩ ﻜل ﺍﺠل ﻭﻤﻥ u0 =
. u u 3 2 = −= 2 19 ﻭ u u 2 1 = 2 15 − = ، u u 1 0 = 2 13 − = (1
v ﻜﻤﺎ ﻴﻠﻲ : 2 n .vn = un −1) ﻨﻌﺭﻑ ﻤﻥ ﺍﺠل ﻜل ﻋﺩﺩ ﻁﺒﻴﻌﻲ nﺍﻟﻤﺘﺘﺎﻟﻴﺔ ( )
(ﺃ 1 1 12 22 nn n n vu u v ( ) ﻭﻤﻨﻪ + + = −= − = n ﺃﺴﺎﺴﻬﺎ ﻫﻨﺩﺴﻴﺔ ﻤﺘﺘﺎﻟﻴﺔ u
ﺍﻷﻭل ﻭﺤﺩﻫﺎ q = 2 0 0 .v u= − =1 1
ﻋﺒﺎﺭﺓ (ﺏ n :n ﺒﺩﻻﻟﺔ v 0 2 n n
n v vq = × =
ﻋﺒﺎﺭﺓ n 12 1 :n ﺒﺩﻻﻟﺔ u n
n n u v= += +
65 (ﺝ n 2 1 65 ﺘﻜﺎﻓﺊ u = n 2 64 ﻭﻤﻨﻪ + = n n = 6 ﺃﻱ =
u ﻭ ﺭﺘﺒﺘﻪ n . 7 ﺇﺫﻥ ﺍﻟﻌﺩﺩ 65 ﺤﺩ ﻤﻥ ﺤﺩﻭﺩ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ( )
(ﺩ 7 7
012 6 0
1 12 ...... 127
1 12
q Sv v v v v
q
⎛ ⎞ − − = ++ + = = = ⎜ ⎟ ⎝ ⎠ − −