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How can you express the following recursion explicitly? \begin{cases} T_0 = 1\\ T_n = 1 + 2\cdot T_{n-1}\\ \end{cases}

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Very similar to…. See my answer there. – lhf May 16 '14 at 13:32
up vote 4 down vote accepted

There is a standard method for solving this kind of thing, which you should easily find if you use the search box.

However in the present case we can calculate the successive values $$1,\,3,\,7,\,15,\,31,\ldots\ ;$$ this is enough to guess that $T(n)=2^{n+1}-1$, which can then be proved by induction.

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Let $T_n=a_n+c$, where $c$ is a constant that will be chosen later. Then our recursion can be rewritten as $$a_n+c=2(a_{n-1}+c)+1.$$ It follows that $$a_n=2a_{n-1} +c+1.$$ If we choose $c=-1$, the recurrence becomes $$a_n=2a_{n-1},$$ with $a_0=2$. Thus $a_n=2^{n+1}$ and therefore $T_n=2^{n+1}-1$.

Remark: The above trick can be used on $T_n=pT_{n-1}+q$, where $p\ne 1$, and analogues can be used in somehat more complicated situations.

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Let $T_n = U_n - 1$ for all $n$. Then the recursion written in terms of $U$ is $$U_n - 1 = 1 + 2(U_{n-1} - 1) = 2 U_{n-1} - 1,$$ or $$U_n = 2 U_{n-1},$$ with initial condition $U_0 = T_0 + 1 = 2$. So $U_n = 2^{n+1}$ by inspection, and the rest is straightforward.

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What's the motivation for $T_n = U_n - 1$? – 200_success May 16 '14 at 17:01

$T_n = 1+2T_{n-1}$. Shift the index. $T_{n+1} = 1+2T_n$. Subtract.

$T_{n+1}-T_n = 2T_n-2T_{n-1}$ or $T_{n+1}-3T_n+2T_{n-1} = 0$.

Now solve it like a general linear recurrence by finding the characteristic polynomial and so on.

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From $T_{n+1}-T_n = 2T_n-2T_{n-1}$ we get $T_{n+1}-T_n = 2^n(T_1-T_0)$. – lhf May 16 '14 at 14:20

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