Number series – Find the wrong number (multiply by n then add n) 3, 4, 10, 32, 136, 685, 4116 Exactly one term is incorrect. The intended rule is a(n) = a(n−1)*n + n (with n = 1, 2, 3, …). Identify the wrong number.

Introduction / Context:
Some sequences use a tidy recurrence a(n) = a(n−1)*n + n. One bad insertion breaks two adjacent gaps; fixing that single value restores the pattern across the run.

Given Data / Assumptions:

• Series: 3, 4, 10, 32, 136, 685, 4116
• Intended rule: a(n) = a(n−1)*n + n (n = 1, 2, 3, …)

Concept / Approach:
Validate each transition against the recurrence and find the outlier.

Step-by-Step Solution:
3 → 4: 3*1 + 1 = 4 → OK.4 → 10: 4*2 + 2 = 10 → OK.10 → 32: expected 10*3 + 3 = 33 → series shows 32 → wrong here.With 33 in place, 33 → 136: 33*4 + 4 = 136 → OK.136 → 685: 136*5 + 5 = 685 → OK.685 → 4116: 685*6 + 6 = 4116 → OK.

Verification / Alternative check:
The single correction 32 → 33 yields a perfect recurrence throughout, proving “32” is the unique wrong number.

Why Other Options Are Wrong:

• 10, 136, 4116 → all comply once the third term is corrected.
• None of these → incorrect because “32” is the only mismatch.

Common Pitfalls:
Assuming multiple errors; well-posed exam items almost always contain exactly one erroneous insertion.

Final Answer:
32