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Find the general term of the sequence defined by:

$a_n = 2a_{n-1} + 1$ where $a_1$ is given

Thank You

user962768
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  • Try writing out the first few terms, then try to connect what you see to powers of 2. – Robert Cardona Feb 09 '13 at 09:27
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    Since you are new to math.stackexchange, I won't downvote your question. But please, read the faq's to learn how to ask questions on this site.

    Specifically, show us your partial work, or at least explain better your doubts about the problem.

     – Andrea Orta Feb 09 '13 at 09:27

2 Answers2

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Consider : $ a_{n}=b_n +c$, the recurrence relation would be :

$b_n+c=2b_{n-1}+2c+1$

$b_n=2b_{n-1}+c+1$

what if $c=-1$ ? notice that we can assume that $c$ equals any number since we always can have a sequence $b_n$ satisfying the first equality : $a_n=b_n+c$.

$b_n=2b_{n-1} \Longrightarrow b_n=b_0 2^n \ \ $ (Geometric sequence)

Hence, $a_n=b_n+c=b_n-1=b_02^n-1$

$\displaystyle a_1=2b_0-1 \Longleftrightarrow b_0= \frac{a_1+1}{2} $

Finally, you get :

$\displaystyle a_n= \left(\frac{a_1+1}{2}\right)2^n-1 $

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$ \begin{align*} a_n &= 2a_{n-1} + 1 = 2(2a_{n-2} + 1) + 1 = 2^2a_{n-2} + (1 + 2) = 2^2(2a_{n-3} + 1) + (2^0 + 2^1) \\ &= 2^3a_{n-3} + (2^0 + 2^1 + 2^2) \\ &= 2^na_0 + (2^0 + 2^1 + \dots + 2^{n-1}) \\ &= 2^na_0 + 2^n -1 \end{align*} $

Without the value of the first term $a_0$, this is the best you can get.

Paresh
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