The problem appears to be asking for the sequence of the number of dandelions each Saturday. The problem gives that on the first Saturday, the number of dandelions is 225. So

n[0] = 225

To get from any Saturday's number to the next Saturdays number, you subtract 100 and double the result. The equation for that is

n[k+1] = 2(n[k]-100)

The above is called a recursion equation, because it tells you how to get the next in the sequence from the last value or values. Crunching the numbers for the first few in the sequence, I get

n[1] = 250

n[2] = 300

n[3] = 400

The real question is, can you find a closed expression for n[k]? That is, if I tell you which member of the sequence I want to know the value of, is there a way, without calculating all the n's up to that point, that you can compute for me the value of that member of the sequence. Look at the following table of expanding the calculation of the first few values, and you can see a pattern:

n[0] = 225

n[1] = 2^{1}(225) - 2^{1}(100)

n[2] = 2^{2}(225) - 2^{2}(100) - 2^{1}(100)

n[3] = 2^{3}(225) - 2^{3}(100) - 2^{2}(100) - 2^{1}(100)

In general, for the k'th value, you take 225 times 2

^{k} and subtract the sum of all the powers of 2 from there down times 100. The sum of powers of 2 from 1 to k is 2

^{k+1} - 2. So the general formula for the k'th value is

n[k] = 2^{k}(225) - 100(2^{k+1} - 2)

I think this is the kind of work the problem was asking for.

Challenge: Can you prove that the above formula is equivalent to

n[k] = 200 + 2^{k}(25)