Question 25 of 27intermediate🔍 AnalyzeLong Answer4 marks

Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.

Correct Answer

Given: Three consecutive numbers: (n−1)(n-1), nn, and (n+1)(n+1)

Square of middle number: n2n^2

Product of other two: (n−1)(n+1)(n-1)(n+1)

Using Identity 1C: (n−1)(n+1)=n2−1(n-1)(n+1) = n^2 - 1

Difference: n2−(n2−1)=1n^2 - (n^2 - 1) = 1

Pattern: The difference is always 1.

Algebraic Equation:

n2−(n−1)(n+1)=1n^2 - (n-1)(n+1) = 1

Verification (Expanding both sides): LHS = n2−(n2−1)n^2 - (n^2 - 1) LHS = n2−n2+1n^2 - n^2 + 1 LHS = 11 = RHS

∴ This is a true identity.

Exercise: Figure it Out (Section 3) | Q: 7 | (Page Page 21)
For More Understanding

Explanation

The chapter encourages discovering patterns through algebra. Three consecutive numbers can be written as n−1n-1, nn, n+1n+1. Using Identity 1C: (n−1)(n+1)=n2−1(n-1)(n+1) = n^2 - 1. So n2−(n2−1)=1n^2 - (n^2 - 1) = 1 always.

Solution Steps

  1. Three consecutive numbers: (n−1)(n-1), nn, (n+1)(n+1)

  2. Square of middle number: n2n^2

  3. Product of other two: (n−1)(n+1)(n-1)(n+1)

  4. Apply Identity 1C: (n−1)(n+1)=n2−1(n-1)(n+1) = n^2 - 1

  5. Calculate difference: n2−(n2−1)=1n^2 - (n^2 - 1) = 1

  6. Write identity: n2−(n−1)(n+1)=1n^2 - (n-1)(n+1) = 1

  7. Verify by expanding LHS to confirm it equals RHS

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