Question 1085106
{{{ a[1] = 2a[0] - 1 }}}
{{{ a[2] = 2a[1] - 1 }}}
…
{{{ a[n] = 2a[n-1] - 1 }}}

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To find the explicit formula, let's just do some substitutions on the first few values to see if it can be generalized:
{{{ a[1] = 2a[0] - 1 }}}
{{{ a[2] = 2a[1] - 1 = 2(2a[0] - 1) - 1 = (2^2)*a[0] - 2 - 1 }}}
{{{ a[3] = 2a[2] - 1 = 2*(2^2*a[0] - 2 - 1) - 1 = (2^3)a[0] - 2*2 - 2 - 1 }}}

Those last three terms look a lot like  {{{ 2^n - 1 }}}  (n=3 and the numbers add to 7)
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I claim that  {{{ a[n] = (2^n)a[0] - (2^n-1) }}}

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Check:    {{{ a[4] = (2^4)(3/2) - (2^4-1) = 24 - 15 = 9 }}}  (ok)