Question

compute \displaystyle \sum _{k=1}^{K} \sum _{t=1}^{T} 0.5^ k=\quad

Answer 1

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 $0.5T - 0.5^KT$.