how do you use a(r^n -1)/ (r-1) to find sum to infinity without just changing the signs to a(1- r^n)/ (1-r)?

Question

how do you use a(r^n -1)/ (r-1) to find sum to infinity without just changing the signs to a(1- r^n)/ (1-r) ?

Our teacher said that a(r^n -1)/ (r-1) is used with the restriction that r>1 while a(1- r^n)/ (1-r) is used when r<1 so we can't use a(r^n -1)/ (r-1) to find sum to infinity so she is wrong?

Answer 1

unfortunately this is a wrong statement.

both the forms can be used as both are same the above is used as convenient and not as rule

when r > 1 r^n > 1 and when n -> inf this is infinite so it diverges

when r < 1 r^n < 1 and when n->0 and this converges

I shall illustrate as n-> inf

a(r^n-1)/(r-1) = a(0-1)/(r-1) = -a/(r-1) = a/(1-r)

a(1-r^n)/(1-r) = a(1-0)/(1-r) = a/(1-r)

Answer 2

Forget all the baloney your teacher's been babbling for a while.
Forget about those restrictions for r and look at both formula carefully, consider one of them and remember from algebra 101 that if you subject both numerator and denominator of a fraction to the same operation the fraction is unchanged given that the operation corresponds to a uniquely mapping function.
What did that all mean? Well consider
S = a(1- r^n)/ (1-r) --------(1)
Now divide the top and bottom of the right hand side by (-1), now you are left with
S = a(1- r^n)/ (1-r) = a(r^n -1)/ (r-1)
which is evident if you consider the most elementary summation of a geometric progression, the negatives have effecively cancelled out. Now back to why the teacher gave you restrictions, first you might notice that in (1) having values of r>1 will give negative values in the top AND the bottom. Since a student may make a mistake with the negatives and assume that a particular series has a negative sum when it does not the restrictions are apparently put in place to restrict errors.

Considering the sum to infinity, note how as n tends to infinity and r<1 (Can't be equal), r^n goes to zero.
Then,
S[infinity] = a(1-0)/(1-r) = a/(1-r)

Hope this helps!

Answer 3

Yes; she is wrong. She probably only said it because it would make things easier. Because it's actually true.

Algebraically, a(r^n - 1) / (r - 1) is the same as a(1 - r^n) / (1 - r).

Proof:

S = a(r^n - 1) / (r - 1)

Let's factor out a minus sign on the top and bottom.

S = a[(-1)(-r^n + 1)] / [(-1)(-r + 1)]

Switching around the terms in the brackets,

S = a[(-1)(1 - r^n)] / [(-1)(1 - r)]

And now, we can cancel the (-1) on the top and bottom.

S = a[(1 - r^n)/(1 - r)]


***
Here's proof of why it's generally recommended.

Suppose a = 5, r = 1/2, and n = 8. Then, if we used the first form of the formula, we would get

S = 5( [1/2]^8 - 1) / (1/2 - 1)
S = 5 (1/256 - 1) / (1/2 - 1)
S = 5 (1/256 - 256/256) / (-1/2)
S = 5 (-255/256) / (-1/2)

And it makes us deal with negative numbers, meaning there will *always* be one extra step in our solution.

Answer 4

I'm sure your teacher is just trying to prevent you from calculating with too many negative quantities.

They are exactly the same thing.

If you used a(r^n - 1) / (r - 1) when r > 1, you will get a positive number divided by a positive number, which is positive.

If r < 1 in the same expression, you will get a negative number divided by a negative number, which is still positive.

It's just more convenient to work with positive numbers.
Maybe you could confirm that with her.

Answer 5

the consumer-friendly ratio is (a million/3) /(a million/2) = (a million/3)(2/a million) = 2/3 r=2/3 There are 4 words n=4 a1 =a million/2 a2 = (a million/2)(2/3) = a million/3 sum = a(a million-r^n)/(a million-r) = (a million/2)(a million-(2/3)^4) /(a million-2/3) = (a million/2)(3/a million) (a million-sixteen/eighty one) = sixty 5/fifty 4

Answer 6

She must be a publik skool teacher. It's easier to remember just one of the formulas and deal with negative numerators and denominators when they arise.

Answer 7

I keep reminding to my kids: no matter your teacher is right or wrong, no matter you like your teacher or not, if I discover you fail in the subject your @ss will suffer not hers.
Yes, you are right, but keep it in hush with the others. OK?