النظام الرقمى الخاص بمتتاليات المربع والمثلث

[بحبك يامصر]

السلام عليكم ورحمه الله وبركاته
قبل أن نبدأ رحلتنا معكم احب أن اشكر اشخاص اعزاء على قلبى افتقدهم واحبهم جدا ولكنى ادعوا لهم بظهر الغيب
السيد حماده سلام
السيد ياسر المصرى
السيد عمروا القرشى
السيد ماركوف
هؤلاء الاربعه هم اساتذه التحليل الرقمى والزمنى فى مصر بلا منازع هم من اجتهدوا سابقا وفتحوا الباب للبحث والاجتهاد وأخرين لا اعلمهم كل التحيه والتقدير لهم
لعلنا نكمل جزء مما تركوا لنا
كنت اقرأ لهم بالمنتديات ولا اشارك احاول فى صمت وفشلت مرات عديده اكثر من عامين فى البحث بين الكتب قديمها وجديدها
عن اسلوب تحديد الاتجاه واختيار المناطق الامنه للبيع والشراء
واخير توصلت لنظام رقمى يتبع متتاليات خاصه بالمربع والمثلث تحدد الاتجاه اولا ثم كلاستر البيع والشراء الأمن ونستطيع ان نعتمد عليه فى بورصات الاسهم وليس الفوركس فقط
الهدف من الموضوع ان ننشر الخير ونساعد الأخوه لوجه الله لا نريد شكرا ولا احسانا ولا مبالغات بل نريد ان نتعاون فقط من أجل الخير فقط ونتعلم
كل الطرق تؤدى الى روما ولكن لعل الله يقربنا الى اقصر الطرق فهو نعم المولى ونعم النصير
الحمد لله رب العالمين

[kaloeg]

منور و بالتوفيق ان شاء الله

[kmalex14]

تسجيل حضور ومتابعة بالتوفيق ان شاء الله

[بحبك يامصر]

التوصيه الاولى

بيع اليورو

11790

الهدف

11756

التعزيز

11810

ادره رأس المال هامه فلنتذكر ذلك جميعا

توصيه سوينج بعقد صغير وسيبوها هاتروح لهدفها مهما حصل اذا اغلق اليورو اليوم دايلى تحت 11800 فالهدف الى 11600 ثم 11400

بالله التوفيق لكم فهو نعم المولى ونعم النصير

[بحبك يامصر]

لمتابعى البورصه المصريه احب ان انوه على الاتى

الاتجاه صاعد على المدى القصير الى 9300_مربع البيع_9500 الاكستنشن الى 9820 هابط على المدى المتوسط الى 8100__مثلث الشراء__7920

البورصه السعوديه

انهت التصحيح عند 7200 وتتجه الى 10800 هدف اول الشراء بين 7700____8100

[janoubi]

بالتوفيق ان شاء الله

[بحبك يامصر]

من يتابع ما يحدث فى اليورو مازال البيع شغال ومن نقطه التعزيز الجباره بفضل الله وسوف تصل للهدف المعلن بل ازيد انها سوف تصل الى 11736 ولكنا نكتفى بالقليل لعل الله يزيد
طيب نجهز شغل الدولار ين الذاهب الى 12400 بحول الله وقوته ونرجعلكم وكذلك نؤكد على المسار الهابط لليورو الى 11600 على فريم الدايلى ثم 11400
نتمنى من الاخوه وخاصه المفكر العبقرى تابلت ان يشاركنا الرحله الماتعه وبالله التوفيق فهو نعم المولى ونعم النصير

[Marlon]

من يتابع ما يحدث فى اليورو مازال البيع شغال ومن نقطه التعزيز الجباره بفضل الله وسوف تصل للهدف المعلن بل ازيد انها سوف تصل الى 11736 ولكنا نكتفى بالقليل لعل الله يزيد
طيب نجهز شغل الدولار ين الذاهب الى 12400 بحول الله وقوته ونرجعلكم وكذلك نؤكد على المسار الهابط لليورو الى 11600 على فريم الدايلى ثم 11400
نتمنى من الاخوه وخاصه المفكر العبقرى تابلت ان يشاركنا الرحله الماتعه وبالله التوفيق فهو نعم المولى ونعم النصير

ياااااه، ده الدكتور تابلت بقاله زمن، ده إنتا من أيام زمن المنتدى الجميل أيام عمالقة الرقمي

[بحبك يامصر]

فعلا كان زمن جميل اخى الكريم ياريت الناس ديه ترجع يطموننا عليهم بس ويرجعوا لمعاملهم تانى احنا موافقين ربنا كبير احساسى انهم راجعين

[بحبك يامصر]

نقلوا الموضوع مع ان الهدف ماكنش توصيات وبس لا كان العلم والحوار واستدعاء العمالقه كما تقول الموضوع اكبر من التوصيات

[Marlon]

نقلوا الموضوع مع ان الهدف ماكنش توصيات وبس لا كان العلم والحوار واستدعاء العمالقه كما تقول الموضوع اكبر من التوصيات

أعتقد أنه لم يتم تفهم الهدف من الموضوع، أعتقد أن من شروط وضع أي صفقة أن يتم إرفاق شارت توضح أسباب الدخول لكن أنا أعلم أنه سوء تفاهم مش أكتر

[poland]

جئت حضرتك بعد انتهاء الحفلة, ولم تكتفي بذلك ..أتيت على عبارة (بلا منازع)..كل هذا لانك ضحيّة ما يسمحو لك بمتابعته في المنتديات, وما حجب اعظم

[allmouafa9]

حبيبي والله كنا قلقين عليك اكتير والحمدالله على السلامة كنت اكتب في موضوع الاخ العاشق لفتح موضوع جديد حتى نبهني الاخ اوسو الى ان الموضوع فتح
معاك ومنتظرين مواضيعك العلمية كالعادة

[allmouafa9]

مافي جديد فينك اخي بحبك يامصر

[allmouafa9]

هل هذا في مجال البحث ؟؟؟؟؟ للتحفيز
Activity:
Session 1

Build this triangular pattern with counters, and ask the students to tell you how many counters were used.

Discuss the methods they used like adding the layers 1 + 2 + 3 + 4 + 5 = 15.
Comment that fifteen is a triangular number since it forms a triangular pattern.
Challenge the students to find how many counters would be needed to make the tenth triangular number, that is a triangular pattern with ten counters on the bottom row.
Allow the students access to counters if need be but encourage them to use mental arithmetic to find a solution. Discuss ways in which they solved the problem.
Suggest that one way to find the number of counters in the five-layer triangle may have been to transform it like this:

So 1 + 2 + 3 + 4 + 5 = (1 + 4) + (2 + 3) + 5
= 3 × 5
= 15
Encourage the students to use the same strategy for the tenth pattern:

(1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6) = 5 × 11
= 55
Challenge the students to make up a list of the triangular numbers for as far as they can, beginning with the known numbers; 1, 3, 6, 10, 15, …, 55, …
Discuss what patterns they can see that allows them to work out the next number without making the figure with counters. Note that the pattern goes:

Challenge the students to find the 100^th triangular number, i.e. the sum of the counting numbers to 100:
1 + 2 + 3 +… + 98 + 99 + 100 = (1 + 100) + (2 + 99) + (3 + 98) + …
= 50 × 101
= 5050
Make this square figure using counters.

Ask the students to work out how many counters, in total, were needed to make the figure.
Tell them that 4 × 4 = 16 is a square number as sixteen counters can be arranged into a square figure.
Challenge them to find other square numbers. Compile a collective list as more square numbers are found. Note that the differences between consecutive square numbers are all odd.

Ask the students to look at the square figures of counters and explain the odd differences.

The number of new counters added on to an existing square is always a multiple of two (black counters) plus one (white counter). So the difference is always an odd number.
Challenge the students to find the 100^th square number which is 100 × 100 = 10 000.
Session Two

Revisit the set of triangular numbers. Ask the students to find the sum of consecutive triangular numbers and organise their results in a systematic way. For example:
1 + 3 = 4 Triangular numbers 1 3 6 10 15 …
3 + 6 = 9 Consecutive sums 4 9 16 25 …
6 + 10 = 16
10 + 15 = 25
Ask the students to describe any pattern they see and record it as a conjecture. For example:
“If you add any two consecutive triangular numbers you always get a square number.”
Suggest to the students that they need to come up with a way to prove that the conjecture is true. Their proofs are likely to be geometric. For example:

This may lead to the generalisation:

Require the students to be more specific in terms of expressing their generalisation.
“Can you predict the size of the square number if you are given the size of triangular numbers? For example, if you know that the first triangular number is the sum of,
1 + 2 + 3 + 4 + 5 + 6, can you say how big the square number will be?”
Use more specific examples like that to encourage students to describe the link between triangular numbers and square numbers. Ask them to express their ideas in words first, such as:
“The first triangular number plus the second triangular number equals the square number that is the equivalent to the second triangular number”
Students might recognise that inventing notation like T
₆
for the sixth triangular number, and S
₇
for the seventh square number (7
²
) can be used to describe the generalisation:
T₆ + T₇ = S_{7 ,} leading to T_n + T_n+1 = S_n+1 = (n + 1)²
Session Three

Use Copymaster 1 to provide the students with a sequence of problems related to square numbers. Allow the students to solve the problems independently or in small co-operative groups. Assemble the class to discuss the common properties of the problems.
Problems one and two both contain geometric patterns in which an odd number of triangle or squares is added to the previous pattern. Consideration of the layers in the triangular pattern shows that one, then three, then five, then seven triangles are added. This is the property of the sequence of square numbers found in lessons one and two. Students should realise that the tenth square number is 100 so the window would have ten layers of triangles.
Similarly, rearranging the staircase patterns shows that they can be reformed into squares. For example, here is a reformation of the 4-high staircase.

Therefore the number of small squares found in a 15-high staircase would be the fifteenth square number (15 x 15 = 225).
The garden problem (Problem Three) also involves square numbers, as students should detect from the 9m² and 49m² square fields. The problem requires connections to the square root, that is the side length that produces a square of given area. So the length of a side in the 49m² plot is 7 metres since (√49 = 7). Comparing the side lengths of the given squares 7 metres and 3 metres suggests that the smallest plot must have a side length of 2 metres (area 4m²) so the top left square has a side length of 7 – 2 = 5 metres (area 25 m²). The other plot areas can be worked out from there as 16m² and 36 m².
In Problem Four students should recognise similar number sequences in the “Dance till You Drop” competition. If someone dances for twenty-four hours the last hour will be worth $47 and they will receive a total of 24² (twenty-four squared) dollars. That is $576.
For problem five students will need to be aware that there are different sized squares involved and this is helpful in systematically finding them all. Consider the 4 x 4 checkerboard.
Counters can be used to mark the centre of each square as shown below.

Sixteen 1 × 1 squares Nine 2 × 2 squares Four 3 × 3 squares
…and the large square itself.
So the 3 × 3 checkerboard has 1 + 4 + 9 + 16 = 30 squares (The sum of the first five square numbers) By similar reasoning the 8 × 8 checkerboard will have 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204 different squares.
Exactly the same reasoning can be applied to problem six which shows successive squares being joined together. Students may recognise that the number of floors below each square join is 1 then 3 then 6 then 10, and so forth. These are the triangular numbers so a fifty-five floor building must contain 10 different square sections since 55 is the tenth triangular number. So the total number of windows will be found by 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385 windows.
Session Four

Give the students copies of Copymaster Two that has problems related to triangular numbers. Allow them to solve the problems individually or in co-operative groups.
After a while gather the class together to share their solution strategies.
In solving problem one the students might note that the welds form matching triangular patterns allowing the triangular numbers to be used an efficient way to get the answer. For example, the welds for four joined squares is twice the third triangular number ( 2 x 6 = 12). For ten square twice the ninth triangular number, forty-five gives the number of welds (90).

Problem two involves a pattern in which the next distance in the spiral is the previous distance plus one metre. Students should recognise that this pattern of differences is found in the triangular numbers. The spiral will have a total distance of the tenth triangular number, fifty-five.
This pattern is less recognisable in problems three to five that involve combinations. In the icecream problem each new flavour creates an extra number of icecreams equal to the previous number of flavours plus one. For example, combining Uncool with the other flavours X, Y, and Z, gives four more possible icecreams. They are XU, YU, ZU, and UU. The next flavour Vex will combine with the others to create five more double scoop icecreams XV, YV, ZV, UV, and VV. So the total number of possible icecreams will be the sum of 1 + 2 + 3 + 4 + 5 + 6 = 21 (the sixth triangular number).
Similar reasoning can be used to solve the other two problems. Each new town creates a number of roads equal to the number of existing towns. Consider the addition of the first new town, Dag, results in three new roads, AD, BD, and CD.

So the number of roads with seven town in total will be the sum of 1 + 2 + 3 + 4 + 5 + 6 = 21 (the sixth triangular number).
Problem Five involves a different context but the same mathematics where each new person added must play games with each of the current players. The number of games that must be played is the sum of 1 + 2 + 3 + …+7 + 8 + 9. This is forty-five, the ninth triangular number.
Session Five

Provide the students with the following sequential patterns. Invite them to find out how many counters will be needed to form the tenth member of each pattern.
Swimming Togs

size one size two size three size four
Town Houses

Plan 1 Plan 2 Plan 3 Plan 4
Note that both patterns involve figures that can be partitioned into triangular and square arrays. The swimming togs pattern is made of two triangular arrays joined together and therefore the number of counters is twice the matching triangular number:

Size	1	2	3	4	5	6	7	8	9	10
Counters	6	12	20	30	42	56	72	90	110	132

The townhouse pattern consists of square (body of the house) and triangular (roof line) patterns being joined.

Plan	1	2	3	4	5	6	7	8	9	10
Body	4	9	16	25	36	49	64	81	100	121
Roof	1	3	6	10	15	21	28	36	45	55
Total	5	12	22	35	51	70	92	117	145	176

As a final problem pose this challenge:
This triangular pattern can be changed into a square by moving only a few counters.What is the smallest number of counters that need to be moved?
Students may realise that the figure has 36 counters that could be used to form a 6 x 6 square pattern.



Superimposing the size of a 6 x 6 counter square over the triangular array reveals that only six counters need to be moved.