The sum of the infinite series (1/4)(1/4 − 1/6) + (1/4)^2[(1/4)^2 − (1/6)^2] + (1/4)^3[(1/4)^3 − (1/6)^3] + ⋯ is equal to:
Answer and solution
Correct answer: B
Solution
Answer: B — 8/345
1. Look at the pattern in each term. The first term is (1/4)² − (1/24), and the following terms follow the same two geometric patterns.
2. The complete series can be separated into:
3. (1/16) + (1/16)² + (1/16)³ + ...
minus
(1/24) + (1/24)² + (1/24)³ + ...
4. The sum of an infinite geometric series is first term ÷ (1 − common ratio).
5. First series:
6. (1/16) ÷ (1 − 1/16) = 1/15
7. Second series:
8. (1/24) ÷ (1 − 1/24) = 1/23
9. Required sum = 1/15 − 1/23
10. Taking the LCM:
11. = (23 − 15)/345
= 8/345
12. Answer: 8/345.
Why this method works:
For the nth term, distribute (1/4)ⁿ across the bracket: it becomes (1/16)ⁿ−(1/24)ⁿ. Both common ratios have absolute value below 1, so both infinite geometric sums converge and may be subtracted.
Independent check and common trap:
The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.
Apply: Each term is 16⁻ⁿ−24⁻ⁿ. Sum=1/(16−1)−1/(24−1)=1/15−1/23=8/345.
Check: The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.
1. Look at the pattern in each term. The first term is (1/4)² − (1/24), and the following terms follow the same two geometric patterns.
2. The complete series can be separated into:
3. (1/16) + (1/16)² + (1/16)³ + ...
minus
(1/24) + (1/24)² + (1/24)³ + ...
4. The sum of an infinite geometric series is first term ÷ (1 − common ratio).
5. First series:
6. (1/16) ÷ (1 − 1/16) = 1/15
7. Second series:
8. (1/24) ÷ (1 − 1/24) = 1/23
9. Required sum = 1/15 − 1/23
10. Taking the LCM:
11. = (23 − 15)/345
= 8/345
12. Answer: 8/345.
Why this method works:
For the nth term, distribute (1/4)ⁿ across the bracket: it becomes (1/16)ⁿ−(1/24)ⁿ. Both common ratios have absolute value below 1, so both infinite geometric sums converge and may be subtracted.
Independent check and common trap:
The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.
G Strategy shortcut
Spot: Algebra – Infinite Series.Apply: Each term is 16⁻ⁿ−24⁻ⁿ. Sum=1/(16−1)−1/(24−1)=1/15−1/23=8/345.
Check: The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.
G Strategy: Spot: Algebra – Infinite Series. Apply: Each term is 16⁻ⁿ−24⁻ⁿ. Sum=1/(16−1)−1/(24−1)=1/15−1/23=8/345. Check: The first term is 1/48≈0.020833, while the full sum is 8/345≈0.023188. The later terms are positive, so a result slightly above the first term is sensible; reversing the subtraction would give an impossible negative result.