Question 1205174
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I assume that an = 2^n-1+1 refers to {{{a[n] = 2^(n-1)+1}}}
To make "n-1" the exponent, surround it in parenthesis.
You should write an = 2^(n-1)+1


Base case:
n = 1
{{{a[n+2] = 3*a[n+1]-2*a[n]}}}
{{{a[1+2] = 3*a[1+1]-2*a[1]}}}
{{{a[3] = 3*a[2]-2*a[1]}}}
{{{a[3] = 3*3-2*2}}}
{{{a[3] = 5}}}
and
{{{a[n] = 2^(n-1)+1}}}
{{{a[1] = 2^(1-1)+1}}}
{{{a[1] = 2}}}
and
{{{a[n] = 2^(n-1)+1}}}
{{{a[2] = 2^(2-1)+1}}}
{{{a[2] = 3}}}
and lastly
{{{a[n] = 2^(n-1)+1}}}
{{{a[3] = 2^(3-1)+1}}}
{{{a[3] = 5}}}
The base case is done.


Inductive step:
Assume that {{{a[k] = 2^(k-1)+1}}} for some integer k such that k > 3.
The goal is to show {{{a[k+1] = 2^(k+1-1)+1 = 2^k + 1}}} based on that assumption.


A bit of scratch work off to the side
{{{a[k] = 2^(k-1)+1}}}
{{{a[k-1] = 2^(k-1-1)+1}}}
{{{a[k-1] = 2^(k-2)+1}}}


Now onto the inductive step
{{{a[n+2] = 3*a[n+1]-2*a[n]}}}


{{{a[k-1+2] = 3*a[k-1+1]-2*a[k-1]}}} Reindexing (replace each n with k-1)


{{{a[k+1] = 3*a[k]-2*a[k-1]}}}


{{{a[k+1] = 3*(2^(k-1)+1)-2*(2^(k-2)+1)}}} Substitution. Use the inductive assumption shown above.


{{{a[k+1] = 3*2^(k-1)+3-2*2^(k-2)-2}}}


{{{a[k+1] = 3*2^(k-1)-2*2^(k-2)+1}}}


{{{a[k+1] = 3*2^k*2^(-1)-2*2^k*2^(-2)+1}}}


{{{a[k+1] = (3/2)*2^k-(1/2)*2^k+1}}}


{{{a[k+1] = (3/2-1/2)*2^k+1}}}


{{{a[k+1] = (1)*2^k+1}}}


{{{a[k+1] = 2^k+1}}}
This wraps up the inductive step and the induction proof is complete.


We have shown that the assumption {{{a[k] = 2^(k-1)+1}}} leads to {{{a[k+1] = 2^k + 1}}}


Therefore, the recursive sequence
{{{system(a[1] = 2, a[2] = 3, a[n+2] = 3*a[n+1]-2*a[n])}}}
has the closed form
{{{a[n] = 2^(n-1)+1}}}
A spreadsheet can be used to generate a list of examples. 
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