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Ann has some sticks that are all of the same length - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 2

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Ann has some sticks that are all of the same length. She arranges them in squares and has made the following 3 rows of patterns: Row 1 ☐ Row 2 ☐☐ Row 3 ☐☐☐ She ... show full transcript

Worked Solution & Example Answer:Ann has some sticks that are all of the same length - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 2

Step 1

Find an expression, in terms of n, for the number of sticks required to make a similar arrangement of n squares in the nth row.

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Answer

To determine the number of sticks required for n squares in the nth row, we first observe the pattern:

  • Row 1 uses 4 sticks for 1 square,
  • Row 2 uses 7 sticks for 2 squares,
  • Row 3 uses 10 sticks for 3 squares.

We notice a consistent increase in the number of sticks. For each square, an increase of 3 sticks occurs in each subsequent row. Thus, we can express the number of sticks required for n squares in the nth row as:

Sn=4+(n−1)imes3=3n+1.S_n = 4 + (n - 1) imes 3 = 3n + 1.

Step 2

Find the total number of sticks Ann uses in making these 10 rows.

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Answer

We now sum the number of sticks used in each of the first 10 rows:

S=extTotal=S1+S2+S3+....+S10S = ext{Total} = S_1 + S_2 + S_3 + .... + S_{10}

Substituting the expression we found:

S=(3imes1+1)+(3imes2+1)+(3imes3+1)+...+(3imes10+1)S = (3 imes 1 + 1) + (3 imes 2 + 1) + (3 imes 3 + 1) + ... + (3 imes 10 + 1)

This reveals a series:

S=(3(1+2+3+...+10))+10S = (3(1 + 2 + 3 + ... + 10)) + 10

Using the formula for the sum of the first n integers, where n=10n = 10:

S=3×10(10+1)2+10=3×55+10=165+10=175.S = 3 \times \frac{10(10 + 1)}{2} + 10 = 3 \times 55 + 10 = 165 + 10 = 175.

Step 3

show that k satisfies $k(3 - 100) + (k + 35) < 0$.

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Answer

We start with the expression:

k(3−100)+(k+35)<0k(3 - 100) + (k + 35) < 0

Simplifying:

k(−97)+k+35<0  ⟹  −96k+35<0  ⟹  96k>35  ⟹  k<3596.k(-97) + k + 35 < 0 \implies -96k + 35 < 0 \implies 96k > 35 \implies k < \frac{35}{96}.

Step 4

Find the value of k.

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Answer

From the previous inequality, we can conclude the value of k must satisfy:

k<3596≈0.3646.k < \frac{35}{96} \approx 0.3646.

Since k must be a whole number of complete rows, we find that the maximum value for k is 0, making the only valid value for k:

k=0.k = 0.

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