What is wrong with this solution?

Problem:

Prove that $\displaystyle \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{ n}{2^n} = 2-\frac{n+2}{2^n}$ for $\displaystyle n=1,2,3,...$

Solution:

Proof by induction

Let the LHS =

$\displaystyle

\sum_{k=1}^{n+1}(\frac{n}{2^n})= \sum_{k=1}^{n}(\frac{n}{2^n})+\sum_{n}^{n+1}(\frac {n}{2^n})

$

RHS is $\displaystyle 2-\frac{n+2}{2^n}$

Here really begins your proof. The above stuff is only algebraic manipulations.
Is it valid for n=1?

LHS = $\displaystyle \frac{1}{2}$

RHS= $\displaystyle 2-\frac{1+2}{2^1} = \frac{1}{2}$ True!

Now let LHS =

$\displaystyle

\sum_{k=n}^{n+1}(\frac{n}{2^n}) = \frac{n+1}{2^{n+1}}

$

Thats what we want our RHS to look like:

Now

Lets look att RHS.

Lets set (n+1) into the formula and subract n that should then equal the LHS hopefully:

$\displaystyle

2-\frac{(n+1)+2}{2^{n+1}}-\left( 2-\frac{n+2}{2^{n}} \right)

$

.

$\displaystyle

2-\frac{(n+1)+2}{2^{n+1}}-\left( 2-\frac{2(n+2)}{2\times 2^{n}} \right)

$

.

$\displaystyle

2-\frac{(n+1)+2}{2^{n+1}}- 2+\frac{2(n+2)}{2\times 2^{n}}

$

.

$\displaystyle

\frac{2n+4}{2\times 2^{n}}-\frac{n+3}{2\times 2^{n}}

$

.

$\displaystyle

\frac{n+1}{2\times 2^{n}}

$

RHS = LHS and by induction this is true for all n $\displaystyle \in \textbf{R}$

Why is this not right?