Answer
$(1^2)(2^1)+(2^2)(2^2)+(3^2)(2^3)+(4^2)(2^4)+(5^2)(2^5)+(6^2)(2^6)+(7^2)(2^7)+(8^2)(2^8)+(9^2)(2^9)+(10^2)(2^{10})$
Work Step by Step
We rewrite the given sum by expanding it: in each term $n^22^n$ substitute the value of $n$, where $n=1,2,3,\dots,10$:
$$\begin{align*}
\sum_{n=1}^{10}n^22^n&=(1^2)(2^1)+(2^2)(2^2)+(3^2)(2^3)+(4^2)(2^4)+(5^2)(2^5)+(6^2)(2^6)+(7^2)(2^7)+(8^2)(2^8)+(9^2)(2^9)+(10^2)(2^{10}).
\end{align*}$$
