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Write an explicit formula for the sequence determined by the recursion formula \[t n = 0.5t _{n-1}+n\] 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 :)

Mathematics
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|dw:1339363260817:dw|
I used this site to help me understand how to solve this: http://hcmop.wordpress.com/2012/04/20/using-characteristic-equation-to-solve-general-linear-recurrence-relations/
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Other answers:

your answers are not correct
yeah I just realized that
yes - the equation I ended up with agrees with the values found by mahmit2012
I can let you know what equation I found if you want - or do you want to try it yourself first?
I know the answer, it's tn= ((n+1)(n+2))/2 but I don't know how to get to it
I haven't done problems like this before but found that document and followed it to get the answer
I actually got:\[t_n=80(\frac{1}{2})^n+2(n-1)\]
yeah but the textbook gave me the answer I posted. And that website uses very complex explanations :$
your formula does not produce the right answers
take n=1 - it produces the value 3
yeah I just checked, you're right it gave me the wrong answer.
if you use the formula I derived then it matches all the values (listed by mahmit2012)
I can try and explain the steps I took to derive this
okay, thanks that would be good
|dw:1339364532207:dw|
ok - first it said arrange the recurrence relation in the form:\[At_n+Bt_{n-1}+Ct_{n-2}+...=f(n)\]so I arranged your relation as:\[t_n-0.5t_{n-1}=n\tag{a}\]it then said the general solution when f(n) is zero is \(t_n=A(0.5)^n\). It then said that to take account of f(n) we need to add in a solution of the form \(Bn+C\). So we substitute this into (a) to get:\[Bn+c-0.5(B(n-1)+C)=n\]which leads to:\[0.5Bn+0.5(B+C)=n\]and if we equate coefficients of LHS and RHS we get:\[B=2\]\[B+C=0\implies C=-B=-2\]so the second part of the equation is of the form:\[2n-2\]we add this to the first part to get the answer:\[t_n=A(0.5)^n+2n-2\]then use the fact that \(t_1=40\) to find A.
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hope it all makes sense?
thanks a lot guys :)
yw :)

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