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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 $3k+5$??

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  • $\begingroup$ Your own answer to the first point is true. Understanding your second point why you don't get $3k+5$ but $k+3$ becomes easier when using brackets $4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k}=4-\left(\frac{2(k+2)}{2^k}-\frac{k+1}{2^k}\right)=4-\left(\frac{2(k+2)-(k+1)}{2^k}\right)=4-\frac{k+3}{2^k}$ $\endgroup$ – flonk Apr 27 '14 at 10:10
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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. $$4-\frac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^k=4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k}=4-\frac{(k+1)+2}{2^{(k+1)-1}}$$.

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    $\begingroup$ A-10 do you mean -1 instead of -2? $\endgroup$ – RDizzl3 Apr 18 '14 at 8:22
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    $\begingroup$ Sorry it should be $-1$. $\endgroup$ – Euna Ko Apr 18 '14 at 8:24
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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 $a + (-b + c)$.

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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:

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

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    $\begingroup$ But still, with the new common denominator, if you add them up shouldn't it be 3k+5? I'm probably have a massive blonde moment... $\endgroup$ – Jim Apr 18 '14 at 8:08
  • $\begingroup$ @Jim There is minus in front of $2(k+2)$. $\endgroup$ – Euna Ko Apr 18 '14 at 8:10

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