Question

1.) Use any formula of your choice to find f(40). Explain why you chose that method and what your answer means. Show your calculations.
2.) Given the sequence of numbers: 5, 6, 8, 11, 15, 20, 26, 33, 41,… Explain whether or not this sequence can be considered a function.

Answer 1

Oh, sorry about that, Bill is able to save $35/week after working part-time and paying his expenses. These two formulas show his weekly savings:
f(1) = 35, f(n) = f(1) + f(n-1) for n > 1
f(n) = 35n

Answer 2

There are two formulae for f(n) here. Lets try to use the first formula to find f(40) and see what happens:
f(n) = f(1) + f(n-1)
=> f(40) =f(1) +f(39)
so to know f(40), we need to know f(1) [which is given f(1) =35] and f(39) [which is not given]
To find f(39), you can use the same formula again
f(39) = f(1)+f(38)
again we need to find f(38)!! and so on this continues.......

Answer 3

The second formula is simple:
f(n) = 35n
f(40) = 35X40

Answer 4

as the differences go up by 1 each time, it is a quadratic function

Answer 5

If you analyse the first formula - it is actually same as the second one and gives the same result
f(40) = f(1) + f(39)
= f(1) + f(1) +f(38)
= f(1) + f(1) + f(1) + f(37)
.............................
= f(1) + f(1) +........40 times
= 40 X f(1) = 40 X 35 (because f(1)=35)

Answer 6

if i am not mistaken, if you start with \((0,5)\) you get
\[y=\frac{1}{2}x^2+\frac{1}{2}x+5\]

Answer 7

The sequence is 5,6,8,11 and so on
I can write the sequence as t1,t2,t3,t4 and so on where these represent the terms of the sequence
t2 -t1 = 1
t3-t2 = 2
t4-t3 =3
...........
tn -tn-1=n-1
Adding up all these
tn - t1 = Sum of 1st n-1 numbers = n(n-1)/2
tn = t1 + n(n-1)/2 = 5 + n(n-1)/2

So function is f(n) = 5 + n(n-1)/2