The sum of the series $1+2\cdot 2+3\cdot 2^2+4\cdot 2^3+5\cdot 2^4+\dots +100\cdot 2^{99}\text{ is }$

Let $\mathrm{S}=1+2\cdot 2+3\cdot 2^2+4\cdot 2^3+5\cdot 2^4 +\dots +100\cdot 2^{99}$

$\therefore 2\mathrm{S}=1\cdot 2+2\cdot 2^2+3\cdot 2^3+\dots +99\cdot 2^{99} +100\cdot 2^{100}$

 Substracting, we get

 $\begin{array}{l}-\mathrm{S}=1+1\cdot 2+1\cdot 2^2+\dots +1\cdot 2^{99}-100\cdot 2^{100}\\ =\left(1+2+2^2+\dots +2^{99}\right)-100\cdot 2^{100}\\ =\frac{1\left(2^{100}-1\right)}{2-1}-100\cdot 2^{100}=2^{100}-1-100\cdot 2^{100}\\ \therefore \mathrm{S}=100\cdot 2^{100}-2^{100}+1=99\cdot 2^{100}+1.\end{array}$