How have they done the algebra here?

Question

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 $3k+5$??

Answer 1

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:

$$4-\frac{(k+1)+2}{2^{k+0}} = 4-\frac{(k+1)+2}{2^{(k+1)-1}}$$

Answer 2

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. $$4-\frac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^k=4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k}=4-\frac{(k+1)+2}{2^{(k+1)-1}}$$.

Answer 3

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 $a + (-b + c)$.