T₀ = 3So we are adding a digit 3 in the front of the previous term , while making the next term.The first digit of t_n+1 is 3 and the last n+1 digits of that are t_n.
In the nth term t_n we have n+1 digits of 3.
The last digit of t₁ is t₀.
the last digit of t₂ = last digit of t₁ = t₀
so last digit of t_n is always t₀.
The last two digits of t₁ = t₁ , as t₁ contains 2 digits
the last two digits of t₂ = last 2 digits of t₁ = t₁
So last two digits of t_n for n >= 1, is t₁.
the last 10 digits of t₉ = 3....3 ten times.
The last 10 digits of t₁₀ = last 10 digits of t₉
The last 10 digits of t_k for k > 10, are = t₉
Hence, the last 10 digits of t_k are all same an equal to t₉ for k >= 10.