An expert question on fraction problems for P6, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P5 and P6 qualifying rounds and finals; harder ones work backwards through fractions, or need clever shortcuts with fractions.
The question
Work out the sum below and write it as a fraction m/n in lowest terms (m and n are whole numbers with no common factor except 1).
💡Hint 1: Work out 1/(2 × 3) − 1/(3 × 4) and compare it with the first term, 1/(2 × 3 × 4).
💡Hint 2: Every term splits: 1/(k × (k + 1) × (k + 2)) = 1/2 × [1/(k × (k + 1)) − 1/((k + 1) × (k + 2))]. When you add them all up, the middle ones cancel.
💡Hint 3: After cancelling, only 1/2 × [1/(2 × 3) − 1/(7 × 8)] is left. Remember to reduce it to lowest terms.
Worked solution
Have a go first, then read the solution step by step.
1.Try the first term: 1/(2 × 3) − 1/(3 × 4) = 2/(2 × 3 × 4), exactly 2 times the first term, so the first term = 1/2 × that difference.
2.Every term works the same way: 1/(k × (k + 1) × (k + 2)) = 1/2 × [1/(k × (k + 1)) − 1/((k + 1) × (k + 2))]. Split all 5 terms, one line each.
3.What each line takes away, the next line adds back, so everything in the middle cancels, leaving the first half of the first line and the second half of the last: the sum = 1/2 × [1/(2 × 3) − 1/(7 × 8)] = 1/2 × (1/6 − 1/56).
4.1/6 − 1/56 = 28/168 − 3/168 = 25/168.
5.The sum = 1/2 × 25/168 = 25/336, in lowest terms.
6.m + n = 25 + 336 = 361.
⚠️Common mistake: forgetting the 1/2, or thinking the first and last terms are what's left after cancelling.
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