Answer
$\sum\limits_{k=3}^{7}{\frac{k}{{{2}^{k}}}}$ is $\frac{119}{128}\text{.}$
Work Step by Step
$\sum\limits_{k=3}^{7}{\frac{k}{{{2}^{k}}}}$
For the sum of the notation:
$\sum\limits_{k=3}^{7}{\frac{k}{{{2}^{k}}}}=\frac{3}{{{2}^{3}}}+\frac{4}{{{2}^{4}}}+\frac{5}{{{2}^{5}}}+\frac{6}{{{2}^{6}}}+\frac{7}{{{2}^{7}}}=\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\frac{7}{128}$
And,
$\begin{align}
& \frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\frac{7}{128}=\frac{16\cdot 3+8\cdot 4+4\cdot 5+2\cdot 6+1\cdot 7}{128} \\
& =\frac{48+32+20+12+7}{128} \\
& =\frac{119}{128}
\end{align}$
Thus, the sum of the sigma notation$\sum\limits_{k=3}^{7}{\frac{k}{{{2}^{k}}}}$ is $\frac{119}{128}\text{.}$
