Free help with homework

Why join Brainly?

  • ask questions about your assignment
  • get answers with explanations
  • find similar questions

Consider the sequence t0 = 3; t1 = 33; t2 = 333 ; t3 = 3333

; : : : dened by t0 = 3 and
tn+1 = 3tn for n 0. What are the last two digits in t3 = 3333
? Can you say what the
last three digits are? Show that the last 10 digits of tk are the same for all k 10.



This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.

1 5 1
The Brain
  • The Brain
  • Helper
Not sure about the answer?
Learn more with Brainly!
Having trouble with your homework?
Get free help!
  • 80% of questions are answered in under 10 minutes
  • Answers come with explanations, so that you can learn
  • Answer quality is ensured by our experts