Consider the discrete-time systems T₁ and T₂ defined as follows:

$\begin{array}{c}\left\{T_{1}x\right\}[n]=x[0]+x[1]+\cdots +x[n]\\ \left\{T_{2}x\right\}[n]=x[0]+\frac{1}{2}x[1]+\cdots +\frac{1}{2^n}x[n] \end{array}$

Which one of the following statements is true?

The correct option is: T₁ is not BIBO stable but T₂ is BIBO stable.

1. Understanding BIBO Stability
A system is defined to be Bounded-Input Bounded-Output (BIBO) stable if every bounded input produces a bounded output. Specifically, if the input $x\left[n\right]$ satisfies:
$\left|x\right[n\left]\right|\le M_x<\infty$
for all $n$, then the output $y\left[n\right]$ must satisfy:
$\left|y\right[n\left]\right|\le M_y<\infty$
for all $n$, where $M_x$ and $M_y$ are finite constants.

2. Analyzing System T₁
The output of the first system is given by:
$\left\{T_{1}x\right\}\left[n\right]=\sum _{k=0}^{n}x\left[k\right]$
To test for BIBO stability, we can choose a bounded input and check if it yields an unbounded output. Let us choose a simple bounded input, the constant unit step sequence:
$x\left[n\right]=1$
for all $n\ge 0$. This input is clearly bounded since $\left|x\right[n\left]\right|\le 1$ for all $n$.
Substituting this input into the equation for T₁ gives:
$\left\{T_{1}x\right\}\left[n\right]=\sum _{k=0}^{n}1=n+1$
As $n\to \infty$, the output $(n+1)\to \infty$ and grows without bound. Because a bounded input has produced an unbounded output, T₁ is not BIBO stable.

3. Analyzing System T₂
The output of the second system is defined as:
$\left\{T_{2}x\right\}\left[n\right]=\sum _{k=0}^n{\left(\frac{1}{2}\right)}^{k}x\left[k\right]$
Let us assume any arbitrary bounded input $x\left[n\right]$ such that $\left|x\right[n\left]\right|\le M<\infty$ for all $n$. We check the absolute value of the output:
$\left|\left\{T_{2}x\right\}\right[n\left]\right|=\left|\sum _{k=0}^n{\left(\frac{1}{2}\right)}^{k}x\right[k\left]\right|$
Applying the triangle inequality:
$\left|\left\{T_{2}x\right\}\right[n\left]\right|\le \sum _{k=0}^n{\left(\frac{1}{2}\right)}^k\left|x\right[k\left]\right|$
Since $\left|x\right[k\left]\right|\le M$ for all $k$:
$\left|\left\{T_{2}x\right\}\right[n\left]\right|\le M\sum _{k=0}^n{\left(\frac{1}{2}\right)}^k$
The summation is a finite geometric series with a sum of:
$\sum _{k=0}^n{\left(\frac{1}{2}\right)}^k=\frac{1-{\left(\frac{1}{2}\right)}^{n+1}}{1-\frac{1}{2}}=2[1-{\left(\frac{1}{2}\right)}^{n+1}]<2$
Thus, the magnitude of the output is strictly bounded:
$\left|\left\{T_{2}x\right\}\right[n\left]\right|<2M$
Because the output is guaranteed to remain bounded ($<2M$) for any bounded input ($\le M$), T₂ is BIBO stable.

Therefore, T₁ is not BIBO stable but T₂ is BIBO stable.