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Cutiepo0
Group Title
Write an explicit formula for the sequence determined by the recursion formula tn=(0.5tn1)+n
t1=40, t2=22, t3=13, t4 = 8.5, t5=6.25
Please explain how to get to the answer rather than just the answer. Thanks :)
 2 years ago
 2 years ago
Cutiepo0 Group Title
Write an explicit formula for the sequence determined by the recursion formula tn=(0.5tn1)+n t1=40, t2=22, t3=13, t4 = 8.5, t5=6.25 Please explain how to get to the answer rather than just the answer. Thanks :)
 2 years ago
 2 years ago

This Question is Closed

anonymoustwo44 Group TitleBest ResponseYou've already chosen the best response.0
is it t_n=.5t_n1+1 or t_n=t_n1+n?
 2 years ago

Cutiepo0 Group TitleBest ResponseYou've already chosen the best response.0
sorry, no it's 0.5tn
 2 years ago

anonymoustwo44 Group TitleBest ResponseYou've already chosen the best response.0
ok so it would probobly help if we plug in some values: we know that .5=1/2 so with this we can work easier. t1=40=t0/2+1 so t0=78 now let's plug in some values for n: t1=t0/2+1 t2=t1/2+2=(t0/2+1)/2+2=t0/4+1/2+2 t3=t2/2+3=(t0/4+1/2+2)/2+3=t0/8+1/4+4 t4=t3/2+4=(t0/8+1/4+4)/2+4=t0/16+1/8+6 t5=t4/2+5=(t0/16+1/8+6)/2+5=t0/32+1/18+8 now as you can see, the first term seems to have pattern t0/2^n. the second term seems to have pattern 1/2^{n1} and the third is just a sequence of even numbers which starts with 0 so the formula for that third term will be 2n2. so we have: tn=t0/2^n+1/2^{n1}+2n2 now since t0=78, tn then becomes: tn=(78/2^n)+(1/2^{n1})+(2n2) rechecking, you'll see that using this formula will also give t1=40, t2=22, t3=13, t4 = 8.5, t5=6.25
 2 years ago
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