1MA0 Higher March 2013 Paper 1 Q21
*21 Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers. (4)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \((n + 1)^2 - n^2\) \(= n^2 + 2n + 1 - n^2 = 2n + 1\) \((n + 1) + n = 2n + 1\) OR \((n + 1)^2 - n^2\) \(= (n + 1 + n)(n + 1 - n)\) \(= (2n + 1)(1) = 2n + 1\) \((n + 1) + n = 2n + 1\) OR \(n^2 - (n + 1)^2 = n^2 - (n^2 + 2n + 1) =\) \(-2n - 1 = -(2n + 1)\) Difference is \(2n + 1\) \((n + 1) + n = 2n + 1\) | proof | 4 | M1 for any two consecutive integers expressed algebraically eg \(n\) and \(n + 1\) M1 (dep on M1) for the difference between the squares of ‘two consecutive integers’ expressed algebraically eg \((n + 1)^2 - n^2\) A1 for correct expansion and simplification of difference of squares, eg \(2n + 1\) C1 (dep on M2A1) for showing statement is correct, eg \(n + n + 1 = 2n + 1\) and \((n + 1)^2 - n^2 = 2n + 1\) from correct supporting algebra |