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
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
You are correct, apologies for the mistake. The answer is indeed n = 5.
To solve this problem, let's start by defining the formula for the sum of the first n terms of an exponential sequence with a common ratio of 2.
The sum of the first n terms, denoted by Sn, can be calculated using the formula Sn = a * (r^n - 1) / (r - 1), where a is the first term and r is the common ratio.
Let's denote the sum of the first n terms by S_n and the sum of 2n terms by S_2n.
According to the problem, the sum of 2n terms (S_2n) is 33 times the sum of n terms (S_n).
Therefore, we have the equation:
S_2n = 33 * S_n
We can substitute the formulas for S_2n and S_n into the equation to solve for n.
a * (r^(2n) - 1) / (r - 1) = 33 * a * (r^n - 1) / (r - 1)
Cancelling out the common terms on both sides, we get:
r^(2n) - 1 = 33 * r^n - 33
Since the common ratio is given as 2, we can substitute r = 2 into the equation:
2^(2n) - 1 = 33 * 2^n - 33
Simplifying the equation further:
4^n - 1 = 33 * 2^n - 33
Next, we can isolate the terms involving 2^n on one side of the equation:
4^n - 33 * 2^n = 1 - 33
4^n - 33 * 2^n = -32
Now, we have a quadratic equation. Let's make a substitution to simplify it:
Let's substitute x = 2^n. The equation becomes:
4^(log2(x)) - 33 * x = -32
Using the property of logarithms, we can rewrite 4^(log2(x)) as (2^2)^(log2(x)) = 2^(2 * log2(x)) = 2^(log2(x^2)) = x^2
Now, the equation becomes a quadratic equation in terms of x:
x^2 - 33 * x = -32
Rearranging the terms:
x^2 - 33 * x + 32 = 0
Factoring the quadratic equation gives us:
(x - 1)(x - 32) = 0
Setting each factor equal to zero and solving for x:
x - 1 = 0 or x - 32 = 0
x = 1 or x = 32
Recall that x = 2^n, so we have two possible solutions for n:
2^n = 1 or 2^n = 32
The first equation, 2^n = 1, gives us n = 0.
The second equation, 2^n = 32, gives us n = 5.
Therefore, there are two possible values of n: n = 0 or n = 5, for which the sum of 2n terms is 33 times the sum of n terms.
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.