What is the 150th digit in the series 2345, 23455, 234555, 2345555, ....?

Concept: To find the k-th digit of a series whose terms grow in length according to a fixed pattern, first compute the cumulative digit count added by each term (via direct summation or a closed-form formula) to identify which term contains the target digit, then read the exact digit off that term's known internal structure.

Application:

1. Each term of the series 2345, 23455, 234555, 2345555, ... is the digits "234" followed by a run of 5s: the nth term has n copies of "5", so its length is n + 3 digits.

2. The total number of digits contributed by the first n terms is 3n + n(n+1)/2 (summing (k+3) for k = 1 to n).

3. For n = 14: 3(14) + 14(15)/2 = 42 + 105 = 147 digits, so the first 14 terms together contribute only 147 digits.

4. Since 147 < 150, the 150th digit lies inside the 15th term, at position 150 − 147 = 3 within that term.

5. The 15th term is "234" followed by fifteen 5s, i.e. its digits in order are 2, 3, 4, 5, 5, 5, .... The 3rd digit of this term is 4.

Cross-check: The cumulative count after all 15 terms is 3(15) + 15(16)/2 = 45 + 120 = 165, so the 15th term spans positions 148 to 165 — position 150 does fall inside it, confirming the term is correctly identified. Reading the 15th term's opening digits directly — 2 (1st), 3 (2nd), 4 (3rd) — confirms the 3rd digit is 4.

Result: The 150th digit of the series is 4.