Solved: please answer all of the questions below and explain it, I will rate the answer thank you!

function [out] = midtermRec(n) %determine function of midtermRec with n param
if le(n,2) % it will check whether the value n is less than or equal to 2, if so then the out value is 1.
out = 1
elseif eq(n,3) %if n is equal to 3 then the out value is 2
out = 2
else           % otherwise then out value is sum of the recursion function
out = midtermRec(n-1)+midtermRec(n-3)
end
end

29.midtermRec(1)
The answer is b.1 . Since the n value is 1. So it is true for if le(n,2) case.

30.midtermRec(4)
The answer is b.3
in this case it is false for both if le(n,2) and elseif eq(n,3) case. So it goes to out = midtermRec(n-1)+midtermRec(n-3).

midtermRec(n-1) => midtermRec(3) => here n is 4 so n-1 is 3.call function again using recursion method.
midtermRec(3) this will return the out value is 2. (elseif eq(n,3) out = 2)

then consider midtermRec(n-3),
midtermRec(n-3) => midtermRec(1) => here n is 4 so n-3 is 1. call function again using recursion method.
midtermRec(1) this will return the out value is 1. (if le(n,2) out = 1)

finally add 2 values like out = midtermRec(n-1)+midtermRec(n-3) => out = 2+1, so out=3

31.midtermRec(5);
The answer is d.2,1,3,1,4

midtermRec(5) in this case for both if le(n,2) and elseif eq(n,3) case. So it goes to out = midtermRec(n-1)+midtermRec(n-3).

midtermRec(n-1) => midtermRec(4) => here n is 5 and n-1 is 4. so it will call again midtermRec(4).
midtermRec(4) => in this case for both if le(n,2) and elseif eq(n,3) case. So it goes to out = midtermRec(n-1)+midtermRec(n-3).
midtermRec(n-1) => midtermRec(3) => here n is 4 so n-1 is 3.call function again using recursion method.
midtermRec(3) this will return the out value is 2. (elseif eq(n,3) out = 2)
so first it will print 2

then consider midtermRec(n-3),
midtermRec(n-3) => midtermRec(1) => here n is 4 so n-3 is 1. call function again using recursion method.
midtermRec(1) this will return the out value is 1. (if le(n,2) out = 1)
so second it will print 1

finally add 2 values like out = midtermRec(n-1)+midtermRec(n-3) => out = 2+1, so out=3
and then 3rd it will print as 3

midtermRec(n-3) => midtermRec(2) => here n is 5 and n-1 is 2.
In this case if le(n,2) is true so the out value is 1.
so 4th it will print as 1.

finally add midtermRec(4) + midtermRec(2) from previous case, so 3+1 => 4. So 5th it will print as 4.

32. The answer is c.function y = myFun(n)
if n==1
y=1
else
y=myFun(n-1)+n^2
end
end
consider myFun(5)
false for if n==1 so it will move to else part. here there is a function call of myFun(n-1)
y = 1
y = 5
y = 14
y = 30
y = 55 hence 55 will be returned

as per equation y(n) = n^2+(n-1)^2+(n-2)^2+…+2^2+1,
if replace n=5
y(5) = 5^2+(5-1)^2+(5-2)^2+…+2^2+1
=> 25+16+9+4+1
it will return 55 as y value.