Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 1 · Chapter companion part
Count shared edges in original figures, express growth algebraically and distinguish an nth figure from a sum of figures.
Class 8 English Version: part C of the chapter 1 companion. Eight teaching sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all three parts and review independent proofs and drawings with a teacher.
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Use drawn unit segments or paper strips, not lit matches. Two adjacent squares share a side: 4 + 3 = 7 separate unit segments, not 8. Each new square in a single row adds three. For n connected squares the count is 3n + 1. State whether figures share an edge or are built separately; the same drawing can otherwise invite different material-count interpretations.
The source uses repeating shapes like digits 4, 6 and a two-square column. In the supplied original models, their counts follow 3n + 1, 5n + 1 and 5n + 2. For the first family, the first part costs 4 and each extra part adds 3. For the other two, the first counts are 6 and 7 while each new part adds 5. The same difference does not imply the same starting value.
If the first term is a and each next term adds d, the nth term is a + (n − 1)d for positive integer positions n. For 7,12,17,... this becomes 7 + 5(n − 1) = 5n + 2. The n − 1 counts the gaps after the first term. Check n = 1 as well as a later value. A term formula gives one position; it does not automatically add all previous terms.
One square house with two equal roof slopes uses six unit segments. A neighbouring house shares a vertical wall and adds five new segments. For n houses in one row, H(n) = 6 + 5(n − 1) = 5n + 1. The fourth row has 21 segments and the twentieth has 101. Roof slopes in the drawing have the same length as the wall segments; roof height is shorter than that length.
For the house pattern, H(20) = 101 counts the twentieth figure alone. To build separate figures 1 through 20, sum 6,11,...,101: 20 × (6 + 101) ÷ 2 = 1070. This total assumes separate builds with no reuse of strips. Extending one row successively reuses existing parts and is a different material question. Always state the assumption.
The source creative figure uses 1,3,5,... joined squares at figure k, so the number of cells is 2k − 1. Each row of c squares uses 3c + 1 segments. Substitute c = 2k − 1 to obtain T(k) = 6k − 2. Figure 15 uses 88; figure 21 uses 124. To sum only figures 4 through 15, count 12 terms with endpoints 22 and 88, giving 660 separate segments.
Let the original linear sequence be a(k) = 4k for k = 1,2,... . Adding each neighbouring pair gives b(k) = a(k) + a(k + 1) = 4k + 4(k + 1) = 8k + 4. Its first term is 12. The term at position k − 2 is b(k − 2) = 8k − 12, valid only when k − 2 ≥ 1. Substituting k − 2 into the new formula differs from adding the original terms at positions k and k + 1.
Build a table with position, count, difference and algebraic rule. The rules 2n + 1, 3n + 1 and 4n + 3 are linear; n² − 1 is not because its differences change. A valid proposed rule matches every supplied case and the structural growth. A single agreement does not prove the formula. In written work, state the meaning of the variable and explain where both the repeated and fixed parts come from.
For seven adjacent squares, count 7 top + 7 bottom + 8 vertical segments = 22. This agrees with 3c + 1 at c=7 and with figure rule 6k−2 at k=4. The two variables count different things.
The source house example has H(50)=251 and separate total 50×(6+251)/2=6425. Those calculations agree with the shared-wall rule. The companion uses additional original positions for practice.
The drawn sequence suggests its stated construction rule. To justify an nth-term count, explain how each new part is added; do not rely only on guessing a number sequence.
The source table mixes linear and non-linear expressions. Check the exponent and first term.
| Expression | n = 1 | n = 2 | n = 10 | n = 100 |
|---|---|---|---|---|
| 2n + 1 | 3 | 5 | 21 | 201 |
| 3n + 1 | 4 | 7 | 31 | 301 |
| n² − 1 | 0 | 3 | 99 | 9999 |
| 4n + 3 | 7 | 11 | 43 | 403 |
n is a positive integer position here. The differences of n² − 1 grow, so that row is not a linear pattern.
There are n − 1 gaps after the first term.
T(n) = a + (n − 1)d
The derivation assumes a constant difference and positions starting at 1.
A shared wall is counted once inside each figure. Separate figures are built without reusing strips.
H(n) = 5n + 1 S(n) = n(5n + 7) ÷ 2
Extending one reusable row is a different material question. State which interpretation the task requires.
Figure k contains c = 2k − 1 adjacent squares.
c = 2k − 1 T(k) = 3c + 1 = 6k − 2
The variable c is not the same as the figure position k; substituting one for the other loses half the pattern.
Define a new sequence before substituting a different index.
a(k) = 4k b(k) = a(k) + a(k+1) = 8k + 4 b(k−2) = 8k − 12
Do not substitute k − 2 into only part of a formula, or confuse a sum of two terms with a sum of all terms.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original connected constructions corresponding to the source families. Diagram coordinates use equal unit lengths; do not count a two-unit vertical as one strip.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original geometry. Roof height is √3/2 of the unit edge, so the slopes remain equal in length. These are paper-strip models; no flames or real match-lighting is needed.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original redrawing of the mathematical construction, not copied source artwork. Counts assume a single row with shared sides and equal unit lengths.
Show your working on paper. Check your answer, then choose Next.
10
4 + 3 + 3 = 10; shared sides count once.
31
3 × 10 + 1 = 31.
42
5 × 8 + 2 = 42.
101
5 × 20 + 1 = 101.
1070
20 × (6 + 101) ÷ 2 = 1070.
88
6 × 15 − 2 = 88.
124
6 × 21 − 2 = 124.
660
There are 12 terms; 12 × (22 + 88) ÷ 2 = 660.
403
4 × 100 + 3 = 403.
9999
10000 − 1 = 9999.
52
7 + 9 × 5 = 52.
12
4 + 8 = 12.
44
8 × 5 + 4 = 44.
167
3 × 55 + 2 = 167.
nonlinear
Its early differences are 3,5,7,..., not constant.
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Start at 4 and add 3 for each of n − 1 extra squares: 4 + 3(n − 1) = 3n + 1. Alternatively count n top, n bottom and n + 1 vertical segments: 2n + n + 1.
Both routes must count each shared unit edge once.
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They begin with 4,6,7 respectively and add 3,5,5 for each new repeated part. Use first + added × (n − 1), then simplify.
Relate the fixed remainder to the actual diagram; do not assume equal difference means equal formula.
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Position 1 already contains a. Reaching position n crosses n − 1 gaps of size d, giving a + (n − 1)d. Using nd adds one extra gap.
Check the first position explicitly.
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One house uses six unit segments. Each extra house shares a vertical wall and adds five, so 6 + 5(n − 1) = 5n + 1.
Show the shared edge and the two roof slopes.
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H(20)=101 is one figure. Separate builds require 6+11+...+101=1070 segments. Reusing parts to extend one row is another assumption and does not require that separate-build total.
Name what each count represents and whether parts are reused.
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The endpoints are 6 and 5n+1 with n terms, so S(n)=n(5n+7)/2. For n=20 this is 20×107/2=1070.
State that this sums an arithmetic pattern of separate figures.
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A row with c squares uses 3c+1 unit segments. Substitute c=2k−1 to get 3(2k−1)+1=6k−2.
Keep figure position k separate from square count c.
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The required positions are 4 through 15, so there are 12 terms. T(4)=22 and T(15)=88; the sum is 12×110/2=660.
Do not sum all fifteen figures or count the interval as eleven terms.
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The rows are 3,5,7,9,11,21,201; 4,7,10,13,16,31,301; 0,3,8,15,24,99,9999; and 7,11,15,19,23,43,403.
Show substitution for at least one large position; n²−1 is not linear.
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b(k)=4k+4(k+1)=8k+4. Then b(k−2)=8(k−2)+4=8k−12, for k≥3.
Define the new sequence and its valid positive position range.
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Test several cases including the first, then derive the count from fixed and repeated parts of the structure. A formula that happens to fit one total can still fail the next figure.
Provide one structural reason and a small-case check.
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n²−1 has changing differences, while 4n+3 increases by 4 at each integer position. The exponent matters; containing a letter does not establish linear growth.
Use actual early differences to support the distinction.
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The next has 7 adjacent squares and 3×7+1=22 segments, agreeing with T(4)=6×4−2.
Human review checks the seven squares and shared-side counting.
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Define figure 1 and a constant number of added non-overlapping unit edges. Derive first+added×(n−1), state n is a positive integer, and check at least three figures.
Different correctly explained constructions are acceptable.
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For example under H(n)=5n+1, n=12 gives H(12)=61. That means the twelfth connected house-row figure has 61 unit segments, not that the first twelve figures together use 61.
Require formula, substitution, arithmetic and a correct statement of what is counted.
Make three paper-strip figures, a position table and a written nth-term derivation. Add a separate-build material estimate.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 1, printed pages 1–11. It does not replace the complete chapter or its practical teaching.
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