Find explicit formulas for the following sequences:
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$a_{n+1} = 3a_n+2$ for$n\ge 0$ ,$a_0=0$ solution
$3x/(1-x)(1-3x)$ $3^n-1$
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$a_{n+1} = \alpha a_n + \beta$ for$n\ge 0$ ,$a_0=0$ solution
$\beta x/(1-x)(1-\alpha x)$ ${\alpha^n-1\over \alpha-1}\beta$
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$a_{n+1} = a_n/3 +1$ for$n\ge 0$ ,$a_0=1$ solution
${3/2\over 1-x}-{1/2\over 1-x/3}$ ${3^{n+1}-1\over 2\cdot 3^n}$
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$a_{n+2} = 2a_{n+1}-a_n$ for$n\ge 0$ ,$a_0=0$ ,$a_1=1$ solution
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$x/(1-x)^2$ ; $n$
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$a_{n+2} = 3a_{n+1}-2a_n+3$ for$n>0$ ,$a_0=1$ ,$a_1=2$ solution
${4\over 1-2x}-{3\over (1-x)^2}$ -
$2^{n+2}-3n-3$
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$a_n = 2a_{n-1}-a_{n-2}+(-1)^n$ for$n>1$ ,$a_0=a_1=1$ solution
${1/2\over (1-x)^2}-{1/4\over 1-x}+{1/4\over 1+x}$ ${2n+3+(-1)^n\over 4}$
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$a_n = 2a_{n-1}-n\cdot(-1)^n$ for$n\ge 1$ ,$a_0=0$ solution
${x/9-2/9\over (1+x)^2}+{2/9\over 1-2x}$ ${2^{n+1}-(3n+2)(-1)^n\over 9}$
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$a_n = 3a_{n-1} + {n\choose 2}$ for$n\ge 1$ ,$a_0=2$ solution
${1\over 8}(19\cdot 3^n-2n(n+2)-3)$
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$a_n = 2a_{n-1}-a_{n-2}-2$ for$n > 1$ ,$a_0=a_{10}=0$ solution
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$n(a_1+1-n)$ , so with$a_{10}$ ,$a_n=n(10-n)$
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$a_n = 4(a_{n-1}-a_{n-2})+(-1)^n$ for$n \ge 2$ ,$a_0=1$ ,$a_1=4$ solution
${1+x+x^2\over (1+x)(1-2x)^2} = {1\over 9}{1\over 1+x} +\left({-5\over 18}\right) {1\over 1-2x} + \left({7\over 6}\right){1\over (1-2x)^2}$ ${1\over 9}(-1)^n-{5\over 18}\cdot 2^n+{7\over 6}(n+1)\cdot 2^n$
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$a_n = -3a_{n-1}+a_{n-2}+3a_{n-3}$ for$n\ge 3$ ,$a_0=20$ ,$a_1=-36$ ,$a_2=60$ solution
$5(-3)^n+18(-1)^n-3$
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$a_n = -3a_{n-1}+a_{n-2}+3a_{n-3}+128n$ for$n\ge 3$ ,$a_0=0$ ,$a_1=0$ ,$a_2=0$ solution
$8n^2+28n-29-11(-3)^n+40(-1)^n$