In the series 9, 81, ?, 6561, 59049, which number should replace the question mark to maintain the pattern?

Introduction / Context:
This number series clearly grows very rapidly, suggesting that it is based on exponentiation rather than simple addition or subtraction. Many exam questions use powers of a small integer like 2, 3, or 5 in such contexts. Recognising powers and their exponents is the key idea here.

Given Data / Assumptions:
- The series given is: 9, 81, ?, 6561, 59049.- All terms are large powers of some base, likely 3 or 9.- We assume the exponents increase in a simple, consistent pattern.

Concept / Approach:
We start by factorising each known term into prime powers. If we see a repeated base with changing exponents, we can examine how these exponents progress. Once we identify the base and the exponent sequence, inserting the missing term is straightforward, as it must follow exactly the same exponential rule.

Step-by-Step Solution:
- Rewrite each term as a power of 3.- 9 = 3^2, 81 = 3^4, 6561 = 3^8, 59049 = 3^10.- The exponents we see are 2, 4, ?, 8, 10.- The pattern of exponents increases by 2 each step: 2, 4, 6, 8, 10.- Therefore, the missing exponent must be 6.- 3^6 = 729.- So, the missing term in the series is 729.

Verification / Alternative check:
- Build the complete exponent sequence: 3^2, 3^4, 3^6, 3^8, 3^10.- This yields exactly: 9, 81, 729, 6561, 59049.- All these values fit perfectly into the original series, confirming that 729 is correct.

Why Other Options Are Wrong:
- 3561 and 4213 are not integer powers of 3 with exponent 6 and do not fit the exponential pattern.- If any of these replaced the missing term, the sequence of exponents would be broken and the growth would become inconsistent.- Thus, only 729 matches the established rule of powers of 3 with exponents forming 2, 4, 6, 8, 10.

Common Pitfalls:
Students may try to apply multiplication by a constant ratio; however, the ratio itself changes, causing confusion. Another error is not recognising familiar powers of 3 and instead attempting lengthy trial and error with arbitrary multipliers. Learning common powers of small integers up to reasonable exponents is very helpful for such questions.

Final Answer:
The number that correctly completes the series is 729, so the correct option is 729.