Let $y \in l_2$ and let $$x_k=\left[C_0+\sum\limits_{j=0}^{k-1}\frac{y_j}{(\lambda+1)^{j+1}}\right](\lambda+1)^k.$$
Does there exits unique constant $C_0$, such that $x \in l_2?$
I need to show the convergence of $$\sum\limits_{k=0}^\infty \left|C_0+\sum\limits_{j=0}^{k-1}\frac{y_j}{(\lambda+1)^{j+1}}\right|^2(\lambda+1)^{2k}$$
But I'm stuck on this step.
Let for simplicity $a=\lambda+1$. We will also say that $y_s =0$ whenever $s<0$ and introduce the sequence $$b_j=\begin{cases}a^j,&j\ge 0\\0,&j<0.\end{cases}$$ Your $x_k$ writes $$x_k = ca^k + \sum_{j=0}^{k-1}y_ja^{k-1-j} = cb_k + \sum_{j\in\Bbb Z}y_jb_{k-1-j} = cb_k + (y\star b)_{k-1},$$ where $\star$ denotes the convolution of sequences. By the well-know inequality
$$\|y\star b,\ell_2\|\le \|y,\ell_2\|\cdot \|b,\ell_1\|.$$
If $|a|<1$, then $b\in \ell_2\cap\ell_1$, therefore $y\star b\in \ell_2$ and $cb\in \ell_2$, hence $x\in \ell_2$.
If $|a|\ge 1$, then the situation is more difficult to analyse and it might require some detailed knowledge of the sequence $y$. Among other things, in the case $y=0$, $|a|\ge 1$ we have $x\notin \ell_2$.