When starting a new unit, a teacher announces that each day after the first, the class must do twice the total number of problems that had been assigned on all previous days. The class works for 6 days and on the 7th day she says that this is our last day of the unit. On that last day, what fraction of the problems do they still have to complete.

@hartnn

Question

When starting a new unit, a teacher announces that each day after the first, the class must do twice the total number of problems that had been assigned on all previous days. The class works for 6 days and on the 7th day she says that this is our last day of the unit. On that last day, what fraction of the problems do they still have to complete.

@hartnn

anonymous · Answer

Would you agree that the number of problems increases like this:

1, 2, 6, 18, 54, 162, 486

calculusxy · Answer

What is the rule that is in this sequence?

anonymous · Answer

2 times the sum of all numbers of questions previously assigned

anonymous · Answer

$$a_n = 2\sum_{i=1}^{n-1}a_i$$

calculusxy · Answer

What is this formula? I have no idea what this means.

anonymous · Answer

suppose the teacher assigns 1 problem on the first day.  how many will the teacher assign on day 2?

calculusxy · Answer

2?

anonymous · Answer

okay, so now how many problems have been assigned?

anonymous · Answer

in total.

calculusxy · Answer

If you are talking about how many problems have been answered in Day 2, then I am going with the total of 2 problems.

anonymous · Answer

no, from the beginning, just like it says in the problem.

calculusxy · Answer

Is the sequence going to be like 1,2,4,16... (Somewhere along with that rule?)

anonymous · Answer

please, just stick to the question i asked... i'm trying to help you see the pattern

calculusxy · Answer

I really don't understand your question, I don't get by what you mean by"
okay, so now how many problems have been assigned?"

anonymous · Answer

1 question assigned the first day, yes?
2 questions assigned the second day, yes?

since the beginning, how many questions have been assigned?

calculusxy · Answer

But we don't know if they really assigned 1 question on the first  day. Wouldn't that be like a variable?

anonymous · Answer

doesn't matter, we can multiply by n

anonymous · Answer

n, 2n, etc.

calculusxy · Answer

Ok would that be like n, 2n, 4n, 16n,and so on? Then can solve for the variable.

anonymous · Answer

no... again: please just answer my question.  i'm trying to help you see the pattern.  you obviously have the wrong pattern stuck in your head.  i suggest you remove that, open your mind and please answer my question.

hartnn · Answer

the pattern
1,2,4,8,16,.... is correct, right ?

anonymous · Answer

no

anonymous · Answer

1, 2, 6, 18, 54, 162, 486

hartnn · Answer

" twice the total number of problems that had been assigned on all previous days"
i am sorry, you're right

anonymous · Answer

no worries!

anonymous · Answer

or, n, 2n, 6n, 18n, 54n, 162n, 486n, ...

hartnn · Answer

yeah, even with n=1, you'll get the same final answer, so just go with 
1,2,6,18...

calculusxy · Answer

Can you give me a relevant formula for this pattern?

anonymous · Answer

also since $a_n=2\sum_{i=1}^{n-1}a_i$ and the total number of problems is $a_n+\sum_{i=1}^{n-1}a_i$ we can conclude that

proportion of problems remaining $$=\frac{a_n}{a_n+\sum_{i=1}^{n-1}a_i}=\frac{2\sum_{i=1}^{n-1}a_i}{2\sum_{i=1}^{n-1}a_i+\sum_{i=1}^{n-1}a_i}=\frac{2\sum_{i=1}^{n-1}a_i}{3\sum_{i=1}^{n-1}a_i}=\frac{2}{3}$$

calculusxy · Answer

Can you please translate this? I am not up to that level yet.

anonymous · Answer

can't seem to find a clever way to give a formula for this one.  it's nested sums and i'm a bit rusty on all of this.

anonymous · Answer

$$a_1 = 1 \Rightarrow a_2 = 2\sum_{i=1}^{1}a_i = 2(a_1) = 2(1) = 2$$
$$\Rightarrow a_3 = 2\sum_{i=1}^{2}a_i = 2(a_1+a_2) = 2(a_1+2a_1) = 2(1+2)=6$$
$$\Rightarrow a_4 = 2\sum_{i=1}^{3}a_i = 2(a_1+a_2+a_3) = 2(a_1+2a_1+2(a_1+2a_1)) = 2(1+2+6)=18$$
etc...

calculusxy · Answer

What do you mean. By (2a-1)...

anonymous · Answer

not $2a-1\text{, but }2\cdot a_1$

calculusxy · Answer

Yeah so what does that mean?

anonymous · Answer

2 times $a_1$.  remember, $a_1 = 1$

calculusxy · Answer

What do all those numbers by the big E mean?

anonymous · Answer

don't worry about it.  what level is your class?

calculusxy · Answer

Um seventh

calculusxy · Answer

Ur's?

anonymous · Answer

okay. you won't be seeing this for a while. have a look if you're really interested... http://www.purplemath.com/modules/series.htm

anonymous · Answer

i have a degree in math

calculusxy · Answer

BA or MA?

anonymous · Answer

BA but did about 35 units towards an MA

calculusxy · Answer

Almost with MA?

anonymous · Answer

yeah

calculusxy · Answer

Thanks for ur help. But how did u get 2/3?

anonymous · Answer

do you understand how to get the sequence?

1, 2, 6, 18, 54, 162, 486

anonymous · Answer

if so, the total number of problems is 1+2+6+18+54+162+486 = 729

486/729 = 2/3