Answer
$\sum_{k=1}^{100} \left(k\cdot2^{k+2}\right)$
Work Step by Step
Writing the given sum using sigma notation means to determine the general term of the sum and the limits for the summation index:
$$1\cdot 2^3+2\cdot 2^4+3\cdot 2^5+4\cdot 2^6+\dots+100\cdot 2^{102}=\sum_{k=m}^n a_k.$$
We look for the pattern in the terms of the sum:
$$\begin{align*}
1\cdot 2^3&=1\cdot 2^{2+1}\\
2\cdot 2^4&=2\cdot 2^{2+2}\\
3\cdot 2^5&=3\cdot 2^{2+3}\\
4\cdot 2^6&=4\cdot 2^{2+4}\\
&\dots\\
100\cdot 2^{102}&=100\cdot 2^{2+100}.
\end{align*}$$
So the term on the $k$th position is written as
$$a_k=k\cdot 2^{k+2} .$$
The summation index $k$ goes from $1$ to $100$.
We conclude that the sum can be written using sigma notation like this:
$$1\cdot 2^3+2\cdot 2^4+3\cdot 2^5+4\cdot 2^6+\dots+100\cdot 2^{102}=\sum_{k=1}^{100} \left(k\cdot2^{k+2}\right).$$
