If A = 2^32, B = 2^31 + 2^30 + 2^29 + ... + 2^0 and C = 3^15 + 3^14 + 3^13 + ... + 3^0, which of the following is the correct order of A, B and C?

Introduction / Context:
This question examines your understanding of powers of numbers and geometric series. You are asked to compare three large quantities defined in terms of powers of 2 and powers of 3. Even though the actual numbers are very large, you can use algebraic formulas and approximations to compare their sizes without computing exact decimal values. This type of comparison is common in aptitude tests to check conceptual reasoning with exponents.

Given Data / Assumptions:

• A is defined as 2^32.
• B is defined as the sum 2^31 + 2^30 + 2^29 + ... + 2^0.
• C is defined as the sum 3^15 + 3^14 + 3^13 + ... + 3^0.
• All numbers are positive real numbers.
• We must determine which inequality among A, B, and C is correct.

Concept / Approach:
The sum of a finite geometric series with first term a and common ratio r is given by S_n = a * (r^n - 1) / (r - 1), when r is not equal to 1. For B, the series is a geometric sequence with first term 1 (that is 2^0) and ratio 2, from power 0 up to power 31. For C, the sequence has first term 1 (that is 3^0) and ratio 3, from power 0 up to power 15. Using the geometric series formula, we can express both B and C in compact form and then compare them to A.

Step-by-Step Solution:
First express B as a geometric sum with a = 1 and r = 2. B = 2^0 + 2^1 + 2^2 + ... + 2^31. For such a series, S_n = (r^n - 1) / (r - 1) with n = 32 and r = 2. Therefore B = (2^32 - 1) / (2 - 1) = 2^32 - 1. Thus A = 2^32 and B = 2^32 - 1, so A is larger than B by exactly 1. Now express C as a geometric sum with a = 1 and r = 3. C = 3^0 + 3^1 + 3^2 + ... + 3^15. Using S_n = (r^n - 1) / (r - 1) with n = 16 and r = 3, we get C = (3^16 - 1) / (3 - 1) = (3^16 - 1) / 2. Clearly 2^32 is enormously larger than 3^16, and B is just one less than A. Also, 3^16 is far smaller than 2^32, so (3^16 - 1) / 2 is much smaller than 2^32 - 1. Therefore A > B and B > C, giving A > B > C.

Verification / Alternative Check:
To gain more intuition, compare sizes roughly. Since 2^10 is 1024, which is about 10^3, we get 2^30 approximately equal to 10^9, and 2^32 is about 4 * 10^9. For 3^5, the value is 243, so 3^10 is about 243^2 which is roughly 59000, and 3^15 is about 59000 * 243 which is around 14 million. Hence 3^16 is about 3 * 14 million or roughly 40 million. Even if we double or triple this to account for the sum in C, we still stay in the order of tens of millions. By contrast, 2^32 is in the order of billions. So 2^32 and 2^32 - 1 are both far larger than any combination of the 3 powers in C, confirming the ordering A > B > C.

Why Other Options Are Wrong:

• Option C > B > A: This would mean C is the largest, but C is based on 3 powers only up to 3^15, which yields much smaller values than powers of 2 up to 2^32.
• Option C > A > B: Again this claims C is largest and that A is larger than B, but although A larger than B is true, C is nowhere near their scale.
• Option A > C > B: This suggests C is between A and B, but B equals 2^32 - 1 while A is 2^32, so there is no room for another number between them when comparing exact values.
• Option B > A > C: This reverses A and B, but from B = 2^32 - 1 it is obvious that A = 2^32 is larger than B.

Common Pitfalls:
Some students may be tempted to approximate 3^n as larger than 2^n due to the bigger base and conclude that C could be comparable or larger, without noticing that the exponents are much smaller for powers of 3. Others might forget or misapply the geometric series formula, leading to incorrect closed forms. Also, failing to recognize that B is almost equal to A (just one less) can cause confusion when judging relative sizes. Careful use of formulas and approximate magnitude comparisons helps avoid such errors.

Final Answer:
The correct order of the three quantities is A > B > C.