proof by maths induction not sure how to prove this:
for all positive intergers prove:
\begin{equation}
1+2(2)+3(2^2)+...+n(2^{n-1})=(n-1)(2^n)+1
\end{equation}
heres my try: 
prove $n=1$ : 
\begin{equation}
1=1
\end{equation}
assume true for $n=k$:
\begin{equation}
1+2(2)+3(2^2)+...+k(2^{k-1})=(k-1)(2^k)+1
\end{equation}
prove for
\begin{equation}
1+2(2)+3(2^2)+...+k(2^{k-1})+(k+1)(2^{k+1-1})=(k+1-1)(2^k+1)+1
\end{equation}
therefore:
\begin{equation}
(k-1)(2^k)+1+(k+1)(2^{k+1-1})=(k+1-1)(2^{k+1})+1
\end{equation}stuck from there onwards maybe understanding this wrongly.

Let $S(n)=\sum_{i=1}^n i2^{i-1}$  Your hypothesis is that $S(n)=(n-1)2^n+1$  As you say, it works for $n=1$.  Now assume it is true for $k$ and evaluate $$S(k+1)=S(k)+(k+1)2^k\\=(k-1)2^k+1+(k+1)2^k\\=2k2^k-2^k+1+2^k\\=[(k+1)-1]2^{k+1}+1$$
