Look at the beginning of the recursive formula:
[tex]\sf f(1)=7,f(n)=f(n-1)[/tex]
It gives us the first value, which is 7. The second part means that for whatever term we want to find, we plug that in for 'n'. Say we want to find the second term, let's plug in 2 for 'n':
[tex]\sf f(2)=f(2-1)=f(1)[/tex]
We are left with f(1), which is given as 7. So the second term is 7 and then we have the leftover blank. We already know that the second term is 13. 13 - 7 = 6. Therefore, what must be missing from this recursive formula is + 6.
[tex]\sf f(n)=f(n-1)+6[/tex]
For the explicit function, we want to know what will give us these same values if we plugged in which term we want for 'n'. So if we plugged in 1 for 'n', we want to get 7, and if we plugged in 2 for 'n' we want to get 13, and so on. I can spot the similarity that all of these are additions of 1 to the multiples of 6. 6 * 1 = 6 + 1 = 7, 6 * 2 = 12 + 1 = 13, and so on. We could just write this as:
[tex]\sf 6n+1[/tex]
If you tried plugging in 1, you'd get 7, if you plugged 2 in, you'd get 13, and so on. So this is our explicit formula, the 2nd blank should be 6, and the 3rd blank should be + 1.
For the fourth blank, f(10) would just be plugging in 10 for 'n' into our formula, let's do this:
[tex]\sf 6n+1[/tex]
[tex]\sf 6(10)+1[/tex]
Multiply:
[tex]\sf 60+1[/tex]
Add:
[tex]\sf 61[/tex]
So the 4th blank should be 61.
