An exponential sequence has a common ratio of 2 . Find the value of n for which the sum of 2n terms is 33 times the sum of n

Answer 1

terms.

Let the first term of the exponential sequence be a. Then the sum of n terms is given by:

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

The sum of 2n terms is:

S_2n = a + ar + ar^2 + ... + ar^(2n-1) = a(1 - r^(2n))/(1 - r)

We are given that:

S_2n = 33S_n

a(1 - r^(2n))/(1 - r) = 33a(1 - r^n)/(1 - r)

Canceling out the common factor of a and rearranging, we get:

1 - r^(2n) = 33(1 - r^n)

Substituting r = 2, we get:

1 - 2^(2n) = 33(1 - 2^n)

Expanding 2^(2n) as (2^n)^2 and simplifying, we get:

(2^n - 1)(2^n + 33) = 0

Since 2^n > 1, the only solution is:

2^n + 33 = 0

This is not possible, so there is no value of n that satisfies the given condition.

Answer 2

AAAaannndd the bot gets it wrong yet again!

We got this far
1 - r^(2n) = 33(1 - r^n)
(1-r^n)(1+r^n) = 33(1-r^n)
1+r^n = 33
r^n = 32
2^n = 32
n = 5

Answer 3

You are correct, apologies for the mistake. The answer is indeed n = 5.