Am I properly simplifying this geometric progression?

You've gotten an answer to your specific question, but I want to explain how to correctly solve the question in a mathematically rigorous manner, and not appealing to handwaving. (Every occurrence of "···" is an instance of lack of rigour.)

Working 1

$T(n) - 2·T(n-1) = -1$.

$2 · ( T(n-1) - 2·T(n-2) ) = 2·-1$.   // We do this to match and cancel the $2·T(n-1)$.

$4 · ( T(n-2) - 2·T(n-3) ) = 4·-1$.   // We do this to match and cancel the $4·T(n-2)$.

...   // Not rigorous! We need to express the pattern rigorously.

Solution 1

Take any integer $n > 1$.

$T(n) - 2·T(n-1) = -1$.

$2^k · ( T(n-k) - 2·T(n-k-1) ) = -2^k$ for every integer $k < n-1$.

$\sum_{k=0}^{n-2} ( 2^k · ( T(n-k) - 2·T(n-k-1) ) ) = \sum_{k=0}^{n-2} -2^k$.

$\sum_{k=0}^{n-2} ( 2^k·T(n-k) - 2^{k+1}·T(n-k-1) ) = \sum_{k=0}^{n-2} -2^k$.

$\sum_{k=0}^{n-2} (2^k·T(n-k)) - \sum_{k=0}^{n-2} (2^{k+1}·T(n-k-1)) = -2^{n-1} + 1$.

$\sum_{k=0}^{n-2} (2^k·T(n-k)) - \sum_{k=1}^{n-1} (2^k·T(n-k)) = -2^{n-1} + 1$.

$2^0·T(n-0) - 2^{n-1}·T(n-(n-1)) = -2^{n-1} + 1$.

$T(n) = 2^{n-1}·T(1) - 2^{n-1} + 1 = 2^n + 1$.

Check that $T(1) = 2^1 + 1$ as well, because the above proof assumed $n > 1$.

Solution 2

Take any integer $n > 1$.

$T(n) - 2·T(n-1) = -1$.

$( T(n) - 2·T(n-1) ) / 2^n = -1/2^n$.

$T(n)/2^n - T(n-1)/2^{n-1} = -1/2^n$.

Take any integer $m > 1$.

$\sum_{n=2}^m ( T(n)/2^n - T(n-1)/2^{n-1} ) = \sum_{n=2}^m -1/2^n$.

$\sum_{n=2}^m (T(n)/2^n) - \sum_{n=2}^m (T(n-1)/2^{n-1}) = -1/2 + 1/2^m$.

$\sum_{n=2}^m (T(n)/2^n) - \sum_{n=1}^{m-1} (T(n)/2^n) = -1/2 + 1/2^m$.

$T(m)/2^m - T(1)/2^1 = -1/2 + 1/2^m$.

$T(m) = 2^m·(T(1)/2-1/2) + 1 = 2^m + 1$.

Check that $T(1) = 2^1+1$ as well.

The left sides of your second and third equations should be $T(n)$, not what you have written. Under "Factoring out a $-1$" it would be better to end with $2^{n-3}+2^{n-2}$ to maintain the pattern of the exponents increasing by $1$. Your progression does end with $2^{n-2}$. That is why the numerator of the sum has $2^{n-1}$. If you look at the formula for the sum of a finite geometric progression, the exponent is $1$ more than the highest term: $$1+r+r^2+\ldots +r^k=\frac {r^{k+1}-1}{r-1}$$ When you add $1$ to $n-2$ you get $n-1$ so your teacher's sum of the progression is correct. Your final answer is also correct.

If you are not sure that your calculation is right you should try to verify your calculations.

The first ten numbers of the first recurrence relation

$$T(n) = 2*T(n-1) - 1$$

are

$$3,5,9,17,33,65,129,257,513,1025$$

ten numbers of the second recurrence relation

$$T(n-1) = 2^2*T(n-2) - 2 - 1$$

are

$$3,9,33,129,513,2049,8193,32769,131073,524289$$

So these are different calculations, so something went wrong.

The sentence "the pattern ... represents this relation" is rather unclear. Give it a precise meaning and try to verify it.