$1.3+2.3^{2}+3.3^{3}+\ldots .+\mathrm{n} .3^{\mathrm{n}}=\frac{(2 \mathrm{n}-1) 3^{\mathrm{a}}+\mathrm{b}}{4}$
Let us put $3=\mathrm{x}$. L.H.S: $S=x+2 x^{2}+3 x^{3}+\ldots .+n \cdot x^{n}$
$x s=x^{2}+2 x^{3}+3 x^{4}+\ldots+n \cdot x^{n+1}$
$(1)-(2) \Rightarrow S-x S=\left(x+2 x^{2}+3 x^{3}+\ldots .+n \cdot x^{n}\right)-\left(x^{2}+2 x^{3}\right.$
$\left.+3 x^{4}+\ldots+n \cdot x^{n+1}\right)$
$\Rightarrow S(1-x)=x+x^{2}+x^{3}+\ldots .+x^{n}-n x^{n+1}$
$\Rightarrow S(1-x)=\frac{x\left(1-x^{n}\right)}{1-x}-n x^{n+1}$
$\Rightarrow S=\left(\frac{1}{x-1}\right)\left(\frac{-x\left(x^{n}-1\right)+n x^{n+1}(x-1)}{x-1}\right)$
Put $\mathrm{x}=3$
$\Rightarrow S=\frac{1}{2}\left(\frac{-3^{n+1}+3+2 n \cdot 3^{n+1}}{2}\right)=\left(\frac{3^{n+1}(2 n-1)+3}{4}\right)$