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T₀ = 3
t_{n+1}=3\ t_n\\

So 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.

t_1=3*10^1+t_0=33\\\\t_{n+1}=3\ t_n=3\ *10^{n+1}+t_n

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.

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