Let the term $S_n$ be the sum of the first $n$ powers of $2$. For instance, $S_3 = 2^0 + 2^1 + 2^2 = 7$. Find the largest possible value of

Hi again anon, why don't you join up and take on an individual identity ?  :)

Anyway,  we have

$$\\S_n=2^n-1\\\\
S_{n-1}=2^{n-1}-1=2*2^n-1$$

So we want the greatest common divisor between these two terms.

It is obvious that for many values of n this will be 1 but do not know how to show that this is always the case.

I give this a try（pls someone check this afeter I done)

Let the term $S_n$ be the sum of the first $n$ powers of $2$. For instance, $S_3 = 2^0 + 2^1 + 2^2 = 7$. Find the largest possible value of the greatest common divisor of two consecutive terms, $S_n$ and $S_{n+1}$, for any $n$.

Sn=2^0+2^1+2^2+.......2^(n-1)

consider the pattern 

2^0=(2^1-1)/2-1

2^0+2^1=(2^2-1)/(2-1)

2^0+2^1+2^2=(2^3-1)/(2-1)

2^0+2^1+2^2+2^3=(2^4-1)/(2-1)

.....

2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)=(2^n-1)/(2-1)

2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)+2^n=[2^(n+1)-1]/(2-1)

S{n}=2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)=(2^n-1)/(2-1)

S{N+1}=2^0+2^1+2^2+2^3.......+2^(n-2)+2^(n-1)+2^n=[2^(n+1)-1]/(2-1)

so the greatest common divisor would be 2-1=1

Hi anon, thanks for that answer   

I have NOT checked you logic, I have only coded it so it is more easily read 

Perhaps you shoulod just check that what i have written is what you intended :)

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I give this a try（pls someone check this afeter I done)

Let the term $$S_n$$ be the sum of the first $n$ powers of $2$. For instance,   $$S_3 = 2^0 + 2^1 + 2^2 = 7$$

Find the largest possible value of the greatest common divisor of two consecutive terms, $S_n$ and $S_{n+1}$, for any $n$.

$$\\Sn=2^0+2^1+2^2+.......2^{n-1}\\\\
$consider the pattern$ \\\\
2^0=\frac{(2^1-1)}{2-1}\\\\
2^0+2^1=\frac{(2^2-1)}{(2-1)}\\\\
2^0+2^1+2^2=\frac{(2^3-1)}{(2-1)}\\\\
2^0+2^1+2^2+2^3=\frac{(2^4-1)}{(2-1)}\\\\\\
2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}=\frac{(2^n-1)}{(2-1)}\\\\
2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}+2^n=\frac{2^{n-1}-1}{(2-1)}\\\\
S{n}=2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}=2^{n}-1\\\\
S{n+1}=2^0+2^1+2^2+2^3.......+2^{n-2}+2^{n-1}+2^n=2^{n+1}-1\\\\
$so the greatest common divisor would be $2-1=1\\\\$$

...Thank you Melody.

I dont really know my anwser whther is right or not.I had ever taught by myself how to do this kind of question before.So,I dont really know.

....I found a mistake,the question said ,"for any $n$."when n=-1,S{n}=?

Hi anon,

n has to be greater than or equal to 1 - that is just assumed :)

It's a geometric series with first term 1 and common ratio 2, so

$$S_{n}=2^{n-1}-1,$$

and

$$S_{n+1}=2^{n}-1.$$

Suppose that both of these are exactly divisible by some integer k (say), then there will be integers say s and t such that

$$s\times k = 2^{n-1}-1, \; \text{ and }\; t\times k = 2^{n}-1.$$

Multiply the first of these equations by 2 and subtract that from the second one and you have

$$(t\times k)-(2\times s\times k) = 1,$$

or,

$$k(t-2s)=1.$$

Since k, t and s are integers, it follows that

$$k = 1 \; (\text{and }\; t - 2s=1).$$

Thanks Bertie, that is really neat.  It is so easy when YOU do it.   

s,k and t must all be positive integers I think.   :/

Nit-picker !

But you're correct, I was careless, I should have stated that k was positive.

Notice that the sum of the first n terms and the next term to be added are relatively prime, since their difference is just 1.  For instance:    1  + 2  + 4  = 7.  And the next term to be added is 8.

But, if A, B are relatively prime, then their sum, A + B, (the next  term in the series), must be relatively prime to B, because their difference is just A. [If A and B weren't relatively prime, there would be some factor, k, that would divide them both and would also divide their sum.]

Thus, each successive term in the series is relatively prime to its successor........

Sorry Chris but that is not obvious to me :(

I do understand your's Bertie, and I just could just not resist nit-picking.

It is a very rare oportunity for me to be able to pick you up one anything and I just had to make the most of it!  

Besides, mathematicians are nit-pickers, it is a part of our trade :))

Thank you both for answering.  For a while I did not think anyone was going to   :)