If the sum of the series
(1/2-1/3)+(1/2²-1/(2 · 3)+1/3²)+(1/2³-1/(2² · 3)+1/(2 · 3²)-1/3³)+
(1/2⁴-1/(2³ · 3)+1/(2² · 3²)-1/(2 · 3³)+1/3⁴)+…
is α/β, where α and β are co-prime, then α+3 β is equal to __________.

Solution

Maths sir · 5 steps

Treat each bracket as aⁿ + aⁿ⁻¹b + … + bⁿ with b=-1/3, then sum two GPs.

1. Spot the pattern. Put a = 1/2 and b = -1/3. The nth bracket is aⁿ + aⁿ⁻¹b + … + bⁿ: the powers of a fall while the powers of b rise, and the alternating signs come from b being negative.

   aⁿ + aⁿ⁻¹b + … + bⁿ = (aⁿ⁺¹ - bⁿ⁺¹)/(a - b)

2. Find a minus b. a - b = 1/2 + 1/3 = 5/6, so the nth bracket is 6/5[(1/2)ⁿ⁺¹ - (-1/3)ⁿ⁺¹]. Check with n = 1: 6/5(1/4 - 1/9) = 1/6, which matches 1/2 - 1/3.
3. Sum the two GPs. Add over n = 1, 2, 3, …. First, Σ (1/2)ⁿ⁺¹ = (1/4)/(1 - 1/2) = 1/2. Second, Σ (-1/3)ⁿ⁺¹ = (1/9)/(1 + 1/3) = 1/12.
4. Combine. The sum is 6/5(1/2 - 1/12) = 6/5 × 5/12 = 1/2.
5. So. α/β = 1/2, so α = 1 and β = 2. Then α + 3β = 1 + 6 = 7.

The common slip: 11. Writing a - b as 1/2 - 1/3 = 1/6 instead of 1/2 + 1/3 makes the sum 5/2, which gives 11 instead of 7.

The idea to keep. When a bracket has powers of one number falling while powers of another rise, it is (aⁿ⁺¹-bⁿ⁺¹)/(a-b). Put any alternating sign inside b, then sum the GPs.

Answer from the NTA final answer key. Solution by PrepWiser.