First Order Linear Difference Equations -- Can't understand how the equation is derived

Question

For the case n > n0 + 1, I can't comprehend how Iterating \begin{equation} y[n] = \lambda y[n-1] \end{equation} yields \begin{equation} y[n] = \left(\lambda\right)^{n-n_0}\frac{b \;}{\lambda} \;\;\; n > n_0. \end{equation}

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Note that they have shown that $y[n_0+1]=b$, after which $y[n]=\lambda y[n-1]=\ldots=\lambda^k y[n-k]$ whence $y[n]=\lambda^{n-(n_0+1)}y[n_0+1]=\lambda^{n-n_0}\frac b\lambda$ for $n\gt n_0$ — Prasun Biswas, Aug 23 '21 at 14:16
@PrasunBiswas And -- In the case \lambda \lt 1, how is it that \begin{equation} \lim_{n \rightarrow \infty} y[n] = y[{\infty}]= 0. \end{equation} — Aarush Aggarwal, Aug 23 '21 at 14:33
Because if $0\lt\lambda\lt 1$, the map $n\mapsto\lambda^n$ is strictly decreasing approaching $0$ asymptotically, ie, $\lim_{n\to\infty}\lambda^n=0$ for $0\lt\lambda\lt 1$ — Prasun Biswas, Aug 23 '21 at 14:35

score 0 · Accepted Answer · answered Aug 23 '21 at 14:20

According to the link you provided we have $y[n_0+1]=b$, thus, for $n>n_0$ we may write $$ y[n]= \lambda y[n-1]= \lambda^2 y[n-2]= \ldots= \lambda^{n-n_0-1} y[n-(n-n_0-1)]= \frac{\lambda^{n-n_0}}{\lambda}y[n_0+1]= (\lambda)^{n-n_0}\frac{b}{\lambda}. $$

First Order Linear Difference Equations -- Can't understand how the equation is derived

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