# geometric series

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Quick question:

An infinite geometric series is such that the sum of all the terms after the nth term is equal to twice the nth term. Show that the sum to infinity of the whole series is 3 times the first term.

The solution is already online

Let the series be a, ar, ar^2, ar^3, ..., ar^(n-1), ...

Let S = a + ar + ar^2 + ... + ar^(n-1) be the sum of the first n terms.

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

The sum of all terms after the nth term is

ar^n + ar^(n+1) + ar(n+2) + ...

= r^n(a + ar + ar^2 + ... + ar^(n-1)) + r^(2n)(a + ar + ar^2 + ... + ar^(n-1) + ...

= S r^n + S r^(2n) + S r^(3n) + ...

By hypothesis

S r^n / (1 - r^n) = 2 a r^(n-1)

( a (1 - r^n) / (1 - r)) (r^n / (1 - r^n)) = 2 a r^(n-1)

a r^n / (1 - r) = 2 a r^(n-1)

r / (1 - r) = 2

r = 2 - 2r

r = 2/3

The sum of the whole series is therefore a / (1 - r) = a / (1/3) = 3a

That is to say, the sum of the whole series is 3 times the first term.

The part I dont understand is highlighted in red.

how did he get from ]= r^n (S + S r^n + S r^(2n) + ...)

= S r^n / (1 - r^n basically

thanks for your help

An infinite geometric series is such that the sum of all the terms after the nth term is equal to twice the nth term. Show that the sum to infinity of the whole series is 3 times the first term.

The solution is already online

Let the series be a, ar, ar^2, ar^3, ..., ar^(n-1), ...

Let S = a + ar + ar^2 + ... + ar^(n-1) be the sum of the first n terms.

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

The sum of all terms after the nth term is

ar^n + ar^(n+1) + ar(n+2) + ...

= r^n(a + ar + ar^2 + ... + ar^(n-1)) + r^(2n)(a + ar + ar^2 + ... + ar^(n-1) + ...

= S r^n + S r^(2n) + S r^(3n) + ...

**= r^n (S + S r^n + S r^(2n) + ...)**

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

By hypothesis

S r^n / (1 - r^n) = 2 a r^(n-1)

( a (1 - r^n) / (1 - r)) (r^n / (1 - r^n)) = 2 a r^(n-1)

a r^n / (1 - r) = 2 a r^(n-1)

r / (1 - r) = 2

r = 2 - 2r

r = 2/3

The sum of the whole series is therefore a / (1 - r) = a / (1/3) = 3a

That is to say, the sum of the whole series is 3 times the first term.

The part I dont understand is highlighted in red.

how did he get from ]= r^n (S + S r^n + S r^(2n) + ...)

= S r^n / (1 - r^n basically

thanks for your help

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(Original post by

Quick question:

An infinite geometric series is such that the sum of all the terms after the nth term is equal to twice the nth term. Show that the sum to infinity of the whole series is 3 times the first term.

The solution is already online

Let the series be a, ar, ar^2, ar^3, ..., ar^(n-1), ...

Let S = a + ar + ar^2 + ... + ar^(n-1) be the sum of the first n terms.

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

The sum of all terms after the nth term is

ar^n + ar^(n+1) + ar(n+2) + ...

= r^n(a + ar + ar^2 + ... + ar^(n-1)) + r^(2n)(a + ar + ar^2 + ... + ar^(n-1) + ...

= S r^n + S r^(2n) + S r^(3n) + ...

By hypothesis

S r^n / (1 - r^n) = 2 a r^(n-1)

( a (1 - r^n) / (1 - r)) (r^n / (1 - r^n)) = 2 a r^(n-1)

a r^n / (1 - r) = 2 a r^(n-1)

r / (1 - r) = 2

r = 2 - 2r

r = 2/3

The sum of the whole series is therefore a / (1 - r) = a / (1/3) = 3a

That is to say, the sum of the whole series is 3 times the first term.

The part I dont understand is highlighted in red.

how did he get from ]= r^n (S + S r^n + S r^(2n) + ...)

= S r^n / (1 - r^n basically

thanks for your help

**bigmansouf**)Quick question:

An infinite geometric series is such that the sum of all the terms after the nth term is equal to twice the nth term. Show that the sum to infinity of the whole series is 3 times the first term.

The solution is already online

Let the series be a, ar, ar^2, ar^3, ..., ar^(n-1), ...

Let S = a + ar + ar^2 + ... + ar^(n-1) be the sum of the first n terms.

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

The sum of all terms after the nth term is

ar^n + ar^(n+1) + ar(n+2) + ...

= r^n(a + ar + ar^2 + ... + ar^(n-1)) + r^(2n)(a + ar + ar^2 + ... + ar^(n-1) + ...

= S r^n + S r^(2n) + S r^(3n) + ...

**= r^n (S + S r^n + S r^(2n) + ...)**

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

By hypothesis

S r^n / (1 - r^n) = 2 a r^(n-1)

( a (1 - r^n) / (1 - r)) (r^n / (1 - r^n)) = 2 a r^(n-1)

a r^n / (1 - r) = 2 a r^(n-1)

r / (1 - r) = 2

r = 2 - 2r

r = 2/3

The sum of the whole series is therefore a / (1 - r) = a / (1/3) = 3a

That is to say, the sum of the whole series is 3 times the first term.

The part I dont understand is highlighted in red.

how did he get from ]= r^n (S + S r^n + S r^(2n) + ...)

= S r^n / (1 - r^n basically

thanks for your help

0

reply

(Original post by

Factor out the S and you are left with the sum of an infinite GP inside the bracket..

**brianeverit**)Factor out the S and you are left with the sum of an infinite GP inside the bracket..

I have a question about this;

the formula for the general sum in a geometric series is

where geometric series

since in the general formula there is already a in the numerator do I just replace the with

sorry i just a bit confused

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