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I am looking trying to follow a solution and am failing to understand the following simplification.

Can someone please break the working out down into the smallest steps to help me understand. How do they get from $(\frac{7}{4})^{n-1} + (\frac{7}{4})^{n-2}$ to that being equal to $(\frac{7}{4})^{n-2}\{\frac{7}{4} + 1\}$ and what is with the curly braces? The rest of the solution I understand.

any help is appreciated.

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kimchi lover
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M.Scherzer
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  • $\left(\frac{7}{4}\right)^{n-1}+\left(\frac{7}{4}\right)^{n-2}=\left(\frac{7}{4}\right)^{n-2}\cdot\frac{7}{4}+\left(\frac{7}{4}\right)^{n-2}=\left(\frac{7}{4}\right)^{n-2}\left(\frac{7}{4}+1\right)$, where the first equality is due to $a^{b+c}=a^ba^c$ applied to $a=\frac{7}{4}$, $b=n-2$, and $c=1$. –  Apr 15 '18 at 13:09
  • https://math.stackexchange.com/questions/2722202/if-f-n-is-the-fibonacci-sequence-show-that-f-n-left-frac-74-rightn-fo – lab bhattacharjee Apr 15 '18 at 13:10
  • Curly braces are just another way to group terms, just like parentheses. Why someone uses curly braces instead of parentheses is a matter of style, there is no change in mathematical content. – Lee Mosher Apr 15 '18 at 13:28
  • thanks a bunch @ jyre that's what I was missing. the excellent explanation with the variables a,b,c was super helpful. – M.Scherzer Apr 15 '18 at 13:29
  • but how does the n-1 magically turn into n-2? – M.Scherzer Apr 15 '18 at 13:37

1 Answers1

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Treat the curly brackets as ordinary brackets and factor out $\left(\frac74\right)^{n-2}$:

$\quad\left(\dfrac74\right)^{n-1} + \left(\dfrac74\right)^{n-2} \\= \left(\dfrac74\right)^{n-2}\times \dfrac74 + \left(\dfrac74\right)^{n-2}\times 1 \\=\left(\dfrac74\right)^{n-2}\times \left(\dfrac74+1\right) \\=\left(\dfrac74\right)^{n-2}\times \left(\dfrac{11}4\right)$

Henry
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