compute \displaystyle \sum _{k=1}^{K} \sum _{t=1}^{T} 0.5^ k=\quad
Let us first evaluate the inner sum:
\begin{align*}
\sum _{t=1}^{T} 0.5^ k &= \underbrace{0.5^k + 0.5^k + \ldots + 0.5^k}_{T \text{ times}} \\
&= T \cdot 0.5^k.
\end{align*}
Now, let us evaluate the outer sum:
\begin{align*}
\sum _{k=1}^{K} T \cdot 0.5^k &= T \cdot \left(0.5^1 + 0.5^2 + \ldots + 0.5^K\right) \\
&= T \cdot \frac{0.5(1 - 0.5^K)}{1 - 0.5} \quad \text{(using the formula for the sum of a geometric series)} \\
&= T \cdot \left(0.5 - 0.5^K\right) \\
&= 0.5T - 0.5^KT.
\end{align*}
Therefore, the sum is equal to $0.5T - 0.5^KT$.
To compute the double summation of \(\sum _{k=1}^{K} \sum _{t=1}^{T} 0.5^k\), we need to evaluate the inner summation first and then sum the results for all values of \(k\) from 1 to \(K\).
Let's start by evaluating the inner summation \(\sum _{t=1}^{T} 0.5^k\). Since \(0.5\) does not depend on the variable \(t\), we can treat it as a constant and pull it outside the summation:
\(\sum _{t=1}^{T} 0.5^k = 0.5^k \sum _{t=1}^{T} 1\)
The summation \(\sum _{t=1}^{T} 1\) simply counts the number of terms from 1 to \(T\), which is \(T\):
\(\sum _{t=1}^{T} 0.5^k = 0.5^k \cdot T\)
Now, we need to compute the outer summation:
\(\sum _{k=1}^{K} 0.5^k \cdot T\)
Since \(0.5^k \cdot T\) does not depend on the variable \(k\), we can treat it as a constant and pull it outside the summation:
\(\sum _{k=1}^{K} 0.5^k \cdot T = 0.5^k \cdot T \cdot \sum _{k=1}^{K} 1\)
Again, the summation \(\sum _{k=1}^{K} 1\) simply counts the number of terms from 1 to \(K\), which is \(K\):
\(\sum _{k=1}^{K} 0.5^k \cdot T = 0.5^k \cdot T \cdot K\)
Therefore, the double summation \(\sum _{k=1}^{K} \sum _{t=1}^{T} 0.5^k\) equals \(0.5^k \cdot T \cdot K\).