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Explanation:

The given series is 1234, 12344, 123444, 1234444, .....

So the number of digits in each term are 4, 5, 6, ... or (3  1), (3  2), (3  3), .....upto n terms = 3n n(n 1)23n n(n 1)2

So 3n n(n 1)2≤2003n n(n 1)2≤200

For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184. And 16 them contains 16  3 = 19 digits.

Now 17 term contains 20 digits and 123444......417times123444......4⏟17times. So last digit is 4 and last two digits are 44.