You find it one number at a time. For (a), for example, find $a_1$ by plugging in $1$ everywhere you see an $n$. So, $a_1 = 7a_{1-1}-5 = 7a_0-5$. You are given that $a_0 = 1$, so $a_1 = 7(1)-5 = 2$. Then plug in $2$ for $n$: $a_2 = 7a_{2-1} - 5 = 7a_1-5 = 7(2) - 5 = 9$. Can you figure out $a_3, a_4, a_5$ from there?
Use the same process for (b) through (d).
That is known as a geometric sum. In general, $\dfrac{a^{n+1}-1}{a-1} = \sum_{k=0}^n a^k$
If you want to see the proof of this, do a google search on proof of the geometric sum. In your case, you have a sum that is similar to the right-hand side of the equation. But, instead of $a^k$ inside the summation, you have $\dfrac{8^k}{5^{k+2}}$. You need to play with it to make it look like the formula. Now, $5^{k+2}=5^k\cdot 5^2$, so you can rewrite the summation:
$\displaystyle \begin{align*} \sum_{k=0}^{40} \dfrac{ 8^k }{ 5^{k+2} } & = \sum_{k=0}^{40} \dfrac{ 1 }{ 5^2 } \cdot \dfrac{ 8^k }{ 5^k } \\ & = \dfrac{ 1 }{ 5^2 } \sum_{k=0}^{40} \left( \dfrac{ 8 }{ 5 } \right)^k \end{align*}$
Then let $a=\dfrac{8}{5}$ and plug it into the geometric sum formula above.
$\displaystyle{\sum_{k=0}^{40}}\dfrac{8^k}{5^{k+2} }=$
$\displaystyle{\sum_{k=0}^{40}}\dfrac{8^k}{5^k \cdot 25}=$
$\dfrac{1}{25}\displaystyle{\sum_{k=0}^{40}}\left( \dfrac{8}{5}\right)^k=$
$\dfrac{1}{25}\dfrac{1-\left(\frac{8}{5}\right)^{41}}{1-\frac{8}{5}}=$
$\dfrac{1}{25}\dfrac{\left(\frac{8}{5}\right)^{41}-1}{\frac{3}{5}}=$
$\dfrac{\left(\frac{8}{5}\right)^{41}-1}{15}$
Hello, michaelgg13!
Perhaps you are confused by the notation?
$\displaystyle \text{Find }a_5\text{ given: }\:a_n \:=\:7a_{n-2} - 5,\;a_0 = 1$
We have a sequence of numbers: .$\displaystyle a_0,\,a_1,\,a_2,\,a_3,\,\hdots$
It begins with $\displaystyle a_0 = 1$.
The succesive terms are given by: .$\displaystyle a_n \:=\:7a_{n-1}-5$
This means: $\displaystyle \text{the }n^{th}\text{ term is "7 times the previous term, minus 5".}$
$\displaystyle \begin{array}{cccccccc}\text{So that:} & a_0 &=& 1 \\ & a_1 &=& 7(1) - 5 &=& 2 \\ & a_2 &=& 7(2)-5 &=& 9 \\ & a_3 &=& 7(9)-5 &=& 58 \\ & a_4 &=& 7(58)-5 &=& 401 \\ & a_5 &=& 7(401)-5 &=& 2802 \end{array}$
Get the idea?
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$\displaystyle \text{FYI: }\,\text{The }n^{th}\text{ term is: }\:a_n \;=\;\tfrac{1}{6}(7^n+5)$