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 (plus and minus take turns) and write it as a fraction m/n in lowest terms (m and n are whole numbers with no common factor except 1).
3 − 5/3 + 7/6 − 9/10 … + 31/120 − 33/136 = ?
What is m + n?
Show the hints (3)
💡Hint 1: Try writing the first term, 3, as two fractions with numerator 2 added. And the second?
💡Hint 2: The denominators are 1 + 2 + … + k = k × (k + 1) ÷ 2: each term = 2 × (2k + 1)/(k × (k + 1)) = 2 × (1/k + 1/(k + 1)).
💡Hint 3: Adding and subtracting by turns, the shared fraction of neighbouring terms cancels, leaving only the very first and very last fractions. Find k in the last term, then work out the two that are left.
Worked solution
Have a go first, then read the solution step by step.
1.The first term 3 = 2 × (1/1 + 1/2). In general each term = 2 × (1/k + 1/(k + 1)), with k = 1, 2, 3, … going up by 1.
2.The last term 33/136 has k = 16: 2 × (1/16 + 1/17). From 1 to 16 there are 16 terms, the last with a "−" in front.
3.Written out: 2 × [(1/1 + 1/2) − (1/2 + 1/3) + (1/3 + 1/4) − …]; each term's second fraction and the next term's first are added and taken away, so they cancel.
4.Only the first and last fractions remain: the sum = 2 × (1/1 − 1/17) = 32/17.
5.m + n = 32 + 17 = 49.
⚠️Common mistake: getting the last sign wrong (term 16 of 16 has "−"), giving 1/1 + 1/17; or thinking 1/1 and 1/16 are what's left.
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