How have they done the algebra here? Proof by induction
\begin{align}&4-\frac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^k\\
=&4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k}
\\
=&4-\frac{(k+1)+2}{2^{(k+1)-1}}
\end{align}
Original image
Can someone explain these steps to me please? Did the $2^{k-1}$ change to $2^k$ by multiplying numerator by 2?? Even so, if you add them when they have the common denominator, shouldn't you get $3k+5$??
 A: First step is just multiplying 2 to $2^{k-1}$ and $(k+2)$. And next step is done by adding $-2(k+2)$ and $k+1$. Then it becomes $-(k+3)$ which is $-((k+1)+2)$.
I think you missed the sign.
So, following is true.
$$4-\frac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^k=4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k}=4-\frac{(k+1)+2}{2^{(k+1)-1}}$$.
A: When you look at the second line notice that they have an extra $2$ in the numerator they multiplied by $1$, which can also be written as $1 = \frac{2}{2}$. So just for completeness:
\begin{align*} &4-\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k \\
& = 4-1\cdot\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k \\
& = 4-\frac{2}{2}\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k\\
&=  4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k} \end{align*}
For the numerator you get $-2(k+2)+(k+1) = -(k+3) = -((k+1)+2)$
So it becomes:
$$4-\frac{(k+1)+2}{2^{k+0}} = 4-\frac{(k+1)+2}{2^{(k+1)-1}}$$
A: You've missed the sign! $a - b + c$ is not $a - (b+c)$, or even $a + (b+c)$.
Instead, it is $a + (c-b)$, or equivalently, $a - (b-c)$ or $a + (-b + c)$.
