# Exam Review 3 UPDATE

UDPATE 2 (11/5): The proof of #10 had a typo.
Near the end, after the line “Subtracting 504 from both sides”, the right side should end with a “-504” rather than “+504”.  Similarly, in the next line, the final parentheses should end with “-56” rather than “+56”.

UPDATE: The answer key has been added to the review sheet – let me know if you find an error!

Hi everyone,

Be aware, there was typo in problem number #9.  The left side of the equation should end in “n(n+2)” instead of “n(n+1)”.  Apologies for this error!

-Prof. Reitz

### 2 responses to “Exam Review 3 UPDATE”

1. Albina Yevdayeva

Hello,
my first question is on number 10.
I don’t understand how from:
4^(3(k+1)) + 512 = 576a you get to 4^(3(k+1) )+ 8 = 9(64a+56)
I know that we should get to 4^(3k) + 8 = 9a
And my second question is on number 11:
How do you get:
F2k + F2k+1 = F2k+2 ???
Thanks beforehand

2. Jonas Reitz

Hi Albina,

Thanks for writing – here goes:

In number 10, I’m not sure if it helps but I had a typo – it should read “-56” instead of “+56”. My steps are as follows.
Start with: $4^{3(k+1)}+512=576a$
Subtract 504 from both sides: $4^{3(k+1)}+512 - 504=576a -504$
and combine the numbers on the left to get: $4^{3(k+1)}+8=576a-504$
Now factor 9 out of the right side: $4^{3(k+1)}+8=9(64a-56)$.

For number 11, the basic definition of the Fibonacci numbers includes a rule saying we can obtain $F_n$ by adding the previous two terms $F_{n-2}+F_{n-1}=F_n$. For example, we have $F_{12}+F_{13}=F_{14}$, or $F_{35}+F_{36}=F_{37}$ — basically, if we add two terms in a row, we will get the next one. So what happens if we add $F_{2k} + F_{2k+1}$? For any natural number $k$ you choose, this expression ends up adding two terms in a row – for example, if $k=3$, then $F_{2k} + F_{2k+1}$ will be $F_6+F_7$ – so it should equal the next term. What is the next term after $F_{2k} + F_{2k+1}$? It’s $F_{2k+2}$. Thus $F_{2k} + F_{2k+1}=F_{2k+2}$.

Let me know if you have any followup questions. Good luck!
-Prof. Reitz