Proof of Terras' method for finding T(k)(x) of Collatz sequence. Terras (via Lagarias) has a method explained here (2.2) for finding the kth term of any Collatz sequence (that is)
$$T(n) = \frac{3n+1}{2}$$ if n is odd and
$$=\frac n2$$ if n is even.
The lambda term makes perfect sense as the factor by which n has been increased or decreased, but what is the p(n) term? What does it signify, and how does one prove that this equation yields $T^k(n)$?
 A: $\rho$ is the contribution of all the +1 terms.

For reference, the relevant equations are:
$$
T^{(k)}(n) = \lambda_k (n) n + \rho_k(n)
$$
$$
\lambda_k(n) = \frac{3^{x_0(n) + \cdots + x_{k-1}(n)}}{2^k}
$$
$$
\rho_k(n) = \sum_{i=0}^{k-1} x_i(n) \frac{3^{x_{i+1}(n)+\cdots+x_{k-1}(n)}}{2^{k-i}}
$$
Where $x_i(n)$ is defined as
$$
T^{(i)}(n) \equiv x_i(n) \mod 2
$$

E.g. in the case of $n=3$, $T^{(1)}(3) = 5$ and $T^{(2)}(3) = 8$, which are broken down as:
$$
T^{(1)}(3) = \frac{3}{2} \times 3 + \frac{1}{2}
$$
$$
T^{(2)}(3) = \frac{9}{4} \times 3 + \left( \frac{3}{4} + \frac{1}{2} \right)
$$

Sketch of inductive step in a proof by induction on $k$:
$$
T^{(k)}(n) = \frac{3^{x_{k-1}(n)}}{2}( T^{(k-1)}(n) + x_{k-1}(n) )
=\\= \frac{3^{x_{k-1}(n)}}{2}(\lambda_{k-1}(n) + \rho_{k-1}(n) + x_{k-1}(n) ) =\\
=\lambda_k(n) + \frac{3^{x_{k-1}(n)}}{2} ( \rho_{k-1}(n) + x_{k-1}(n) )
$$
Giving
$$
\rho_k(x) = \frac{3^{x_{k-1}(n)}}{2}( x_{k-1}(n) + \rho_{k-1}(n) ) = \\ =
x_{k-1}(n) \frac{3^{x_{k-1}(n)}}{2} + \sum_{i=0}^{k-2} x_i(n) \frac{3^{x_{i+1}(n)+\cdots+x_{k-1}(n)}}{2^{k-i}} = \\
= \sum_{i=0}^{k-1} x_i(n) \frac{3^{x_{i+1}(n)+\cdots+x_{k-1}(n)}}{2^{k-i}}
$$
